THE EQUAL VARIABLE METHOD

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Equal Variable Method Vasile Cîrtoaje vol. 8, iss. 1, art. 15, 2007

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THE EQUAL VARIABLE METHOD

VASILE CÎRTOAJE

Department of Automation and Computers University of Ploie¸sti

Bucure¸sti 39, Romania.

EMail:vcirtoaje@upg-ploiesti.ro

Received: 01 March, 2006

Accepted: 17 April, 2006

Communicated by: P.S. Bullen 2000 AMS Sub. Class.: 26D10, 26D20.

Key words: Symmetric inequalities, Power means, EV-Theorem.

Abstract: The Equal Variable Method (called alson−1Equal Variable Method on the Mathlinks Site - Inequalities Forum) can be used to prove some difficult symmetric inequalities involving either three power means or, more general, two power means and an expression of formf(x1) +f(x2) +· · ·+f(xn).

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1. Statement of results

In order to state and prove the Equal Variable Theorem (EV-Theorem) we require the following lemma and proposition.

Lemma 1.1. Leta, b, cbe fixed non-negative real numbers, not all equal and at most one of them equal to zero, and letx≤y≤zbe non-negative real numbers such that

x+y+z =a+b+c, x^p+y^p+z^p =a^p+b^p+c^p,

wherep ∈ (−∞,0]∪(1,∞). For p = 0, the second equation isxyz = abc > 0.

Then, there exist two non-negative real numbersx₁ andx₂ withx₁ < x₂ such that x∈[x₁, x₂]. Moreover,

1. ifx=x₁ andp≤0, then0< x < y=z;

2. ifx=x₁ andp > 1, then either0 =x < y ≤z or0< x < y=z;

3. ifx∈(x1, x2), thenx < y < z;

4. ifx=x₂, thenx=y < z.

Proposition 1.2. Leta, b, cbe fixed non-negative real numbers, not all equal and at most one of them equal to zero, and let0≤x≤y ≤zsuch that

x+y+z =a+b+c, x^p+y^p+z^p =a^p+b^p+c^p,

wherep∈(−∞,0]∪(1,∞). Forp= 0, the second equation isxyz =abc >0. Let f(u)be a differentiable function on (0,∞), such that g(x) = f⁰

x^p−1¹

is strictly convex on(0,∞), and let

F₃(x, y, z) = f(x) +f(y) +f(z).

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1. Ifp ≤0, thenF₃ is maximal only for0< x = y < z, and is minimal only for 0< x < y=z;

2. Ifp > 1and eitherf(u)is continuous atu = 0or lim

u→0f(u) = −∞, thenF₃ is maximal only for 0 < x = y < z, and is minimal only for eitherx = 0 or 0< x < y=z.

Theorem 1.3 (Equal Variable Theorem (EV-Theorem)). Leta₁, a₂, . . . , a_n(n ≥ 3)be fixed non-negative real numbers, and let0≤x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂ +· · ·+x_n=a₁+a₂+· · ·+a_n, x^p₁+x^p₂ +· · ·+x^p_n=a^p₁+a^p₂+· · ·+a^p_n,

wherep is a real number,p 6= 1. Forp = 0, the second equation isx₁x₂· · ·x_n = a₁a₂· · ·a_n>0. Letf(u)be a differentiable function on(0,∞)such that

g(x) = f⁰

x^p−1¹

is strictly convex on(0,∞), and let

F_n(x₁, x₂, . . . , x_n) = f(x₁) +f(x₂) +· · ·+f(x_n).

1. If p ≤ 0, then F_n is maximal for 0 < x₁ = x₂ = · · · = xn−1 ≤ x_n, and is minimal for0< x₁ ≤x₂ =x₃ =· · ·=x_n;

2. If p >0and either f(u)is continuous atu = 0 or lim

u→0f(u) =−∞, thenF_n is maximal for 0 ≤ x1 = x2 = · · · = xn−1 ≤ xn, and is minimal for either x1 = 0or0< x1 ≤x2 =x3 =· · ·=xn.

Remark 1. Let 0 < α < β. If the function f is differentiable on (α, β) and the functiong(x) = f⁰

x^p−1¹

is strictly convex on (α^p−1, β^p−1)or(β^p−1, α^p−1), then the EV-Theorem holds true forx1, x2, . . . , xn∈(α, β).

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By Theorem1.3, we easily obtain some particular results, which are very useful in applications.

Corollary 1.4. Let a₁, a₂, . . . , a_n (n ≥ 3) be fixed non-negative numbers, and let 0≤x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂ +· · ·+x_n=a₁+a₂+· · ·+a_n, x²₁+x²₂ +· · ·+x²_n=a²₁+a²₂+· · ·+a²_n.

Letf be a differentiable function on(0,∞)such thatg(x) =f⁰(x)is strictly convex on(0,∞). Moreover, eitherf(x)is continuous atx= 0or lim

x→0f(x) = −∞. Then, Fn =f(x1) +f(x2) +· · ·+f(xn)

is maximal for0 ≤ x₁ = x₂ =· · · = xn−1 ≤ x_n, and is minimal for eitherx₁ = 0 or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Corollary 1.5. Leta₁, a₂, . . . , a_n (n ≥ 3)be fixed positive numbers, and let 0 <

x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂ +· · ·+x_n=a₁+a₂+· · ·+a_n, 1

x₁ + 1

x₂ +· · ·+ 1 x_n = 1

a₁ + 1

a₂ +· · ·+ 1 a_n. Let f be a differentiable function on (0,∞) such that g(x) = f⁰

√1 x

is strictly convex on(0,∞). Then,

F_n =f(x₁) +f(x₂) +· · ·+f(x_n)

is maximal for0< x₁ =x₂ =· · ·=xn−1 ≤ x_n, and is minimal for0< x₁ ≤x₂ = x3 =· · ·=xn.

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Corollary 1.6. Leta₁, a₂, . . . , a_n (n ≥ 3)be fixed positive numbers, and let 0 <

x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂+· · ·+x_n=a₁+a₂+· · ·+a_n, x₁x₂· · ·x_n=a₁a₁· · ·a_n. Letf be a differentiable function on(0,∞)such thatg(x) =f⁰ ¹_x

is strictly convex on(0,∞). Then,

F_n =f(x₁) +f(x₂) +· · ·+f(x_n)

is maximal for0< x₁ =x₂ =· · ·=xn−1 ≤ x_n, and is minimal for0< x₁ ≤x₂ = x₃ =· · ·=x_n.

Corollary 1.7. Let a₁, a₂, . . . , a_n (n ≥ 3) be fixed non-negative numbers, and let 0≤x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂ +· · ·+x_n=a₁+a₂+· · ·+a_n, x^p₁+x^p₂ +· · ·+x^p_n=a^p₁+a^p₂+· · ·+a^p_n, wherepis a real number,p6= 0andp6= 1.

(a) Forp <0,P =x₁x₂· · ·x_nis minimal when0< x₁ =x₂ =· · ·=xn−1 ≤x_n, and is maximal when0< x₁ ≤x₂ =x₃ =· · ·=x_n.

(b) Forp > 0,P =x₁x₂· · ·x_nis maximal when0≤x₁ =x₂ =· · ·=xn−1 ≤x_n, and is minimal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Corollary 1.8. Let a₁, a₂, . . . , a_n (n ≥ 3)be fixed non-negative numbers, let 0 ≤ x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂ +· · ·+x_n=a₁+a₂+· · ·+a_n, x^p₁+x^p₂ +· · ·+x^p_n=a^p₁+a^p₂+· · ·+a^p_n, and letE =x^q₁+x^q₂+· · ·+x^q_n.

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Case 1. p≤0 (p= 0yieldsx₁x₂· · ·x_n=a₁a₂· · ·a_n>0).

(a) Forq ∈(p,0)∪(1,∞),E is maximal when0< x₁ =x₂ =· · ·=xn−1 ≤x_n, and is minimal when0< x₁ ≤x₂ =x₃ =· · ·=x_n.

(b) Forq∈(−∞, p)∪(0,1),Eis minimal when0< x₁ =x₂ =· · ·=xn−1 ≤x_n, and is maximal when0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Case 2. 0< p <1.

(a) Forq ∈(0, p)∪(1,∞),E is maximal when0≤x₁ =x₂ =· · ·=xn−1 ≤x_n, and is minimal when eitherx1 = 0or0< x1 ≤x2 =x3 =· · ·=xn.

(b) Forq∈(−∞,0)∪(p,1),Eis minimal when0≤x1 =x2 =· · ·=xn−1 ≤xn, and is maximal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Case 3. p > 1.

(a) Forq ∈(0,1)∪(p,∞),E is maximal when0≤x₁ =x₂ =· · ·=xn−1 ≤x_n, and is minimal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

(b) Forq∈(−∞,0)∪(1, p),Eis minimal when0≤x₁ =x₂ =· · ·=xn−1 ≤x_n, and is maximal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

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2. Proofs

Proof of Lemma1.1. Leta ≤b ≤ c. Note that in the excluded casesa=b =cand a=b= 0, there is a single triple(x, y, z)which verifies the conditions

x+y+z =a+b+c and x^p+y^p+z^p =a^p+b^p+c^p. Consider now three cases: p= 0,p <0andp >1.

A. Casep= 0 (xyz =abc > 0). LetS = ^a+b+c₃ andP =√³

abc, whereS > P >0 by AM-GM Inequality. We have

x+y+z = 3S, xyz =P³,

and from0< x≤y≤zandx < z, it follows that0< x < P. Now let f =y+z−2√

yz.

It is clear thatf ≥ 0, with equality if and only ify = z. Writingf as a function of x,

f(x) = 3S−x−2P rP

x, we have

f⁰(x) = P x

rP

x −1>0,

and hence the functionf(x)is strictly increasing. Sincef(P) = 3(S−P)>0, the equationf(x) = 0 has a unique positive rootx1, 0 < x1 < P. From f(x) ≥ 0, it follows thatx≥x1.

Sub-case x = x₁. Since f(x) = f(x₁) = 0 and f = 0 implies y = z, we have 0< x < y=z.

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Sub-casex > x₁. We havef(x) >0andy < z. Consider now thatyandz depend onx. Fromx+y(x) +z(x) = 3Sandx·y(x)·z(x) =P³, we get1 +y⁰ +z⁰ = 0 and ¹_x +^y_y⁰ +^z_z⁰ = 0. Hence,

y⁰(x) = y(x−z)

x(z−y), z⁰(x) = z(y−x) x(z−y).

Sincey⁰(x)<0, the functiony(x)is strictly decreasing. Sincey(x₁)> x₁ (see sub- casex =x₁), there existsx₂ > x₁ such thaty(x₂) =x₂,y(x)> xforx₁ < x < x₂ and y(x) < xfor x > x₂. Taking into account that y ≥ x, it follows that x₁ <

x ≤ x₂. On the other hand, we see thatz⁰(x) >0forx₁ < x < x₂. Consequently, the function z(x)is strictly increasing, and hencez(x) > z(x₁) = y(x₁) > y(x).

Finally, we conclude thatx < y < zforx∈(x₁, x₂), andx=y < zforx=x₂. B. Casep < 0. DenoteS= ^a+b+c₃ andR= ^a^p^+b₃^p^+c^p¹_p

. Taking into account that x+y+z = 3S, x^p+y^p+z^p = 3R^p,

from0< x≤y ≤zandx < zwe getx < S and3¹^pR < x < R. Let h= (y+z)

y^p +z^p 2

⁻¹_p

−2.

By the AM-GM Inequality, we have h≥2√

yz 1

√yz −2 = 0,

with equality if and only ify=z. Writing nowhas a function ofx, h(x) = (3S−x)

3R^p−x^p 2

⁻¹_p

−2,

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from

h⁰(x) = 3R^p 2

3R^p −x^p 2

^−1−p_p "

S x

R x

−p

−1

#

>0 it follows thath(x)is strictly increasing. Sinceh(x) ≥ 0andh

3^p¹R

= −2, the equationh(x) = 0has a unique rootx₁ andx≥x₁ >3¹^pR.

Sub-casex = x₁. Since f(x) = f(x₁) = 0, andf = 0implies y = z, we have 0< x < y=z.

Sub-casex > x1. We haveh(x)>0andy < z. Consider now that yandz depend onx. Fromx+y(x)+z(x) = 3Sandx^p+y(x)^p+z(x)^p = 3R^p, we get1+y⁰+z⁰ = 0 andx^p−1 +y^p−1y⁰+z^p−1z⁰ = 0, and hence

y⁰(x) = x^p−1−z^p−1

z^p−1−y^p−1, z⁰(x) = x^p−1−y^p−1 y^p−1−z^p−1.

Sincey⁰(x)>0, the functiony(x)is strictly decreasing. Sincey(x₁)> x₁ (see sub- casex=x₁), there existsx₂ > x₁ such thaty(x₂) =x₂,y(x)> xforx₁ < x < x₂, and y(x) < x for x > x₂. The condition y ≥ x yields x₁ < x ≤ x₂. We see now that z⁰(x) > 0for x₁ < x < x₂. Consequently, the function z(x) is strictly increasing, and hencez(x) > z(x₁) = y(x₁) > y(x). Finally, we havex < y < z forx∈(x₁, x₂)andx=y < zforx=x₂.

C. Casep >1. DenotingS= ^a+b+c₃ andR= ^a^p^+b₃^p^+c^p¹_p yields x+y+z = 3S, x^p+y^p+z^p = 3R^p.

By Jensen’s inequality applied to the convex functiong(u) = u^p, we haveR > S,

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and hencex < S < R. Let

h= 2

y+z

y^p+z^p 2

¹_p

−1.

By Jensen’s Inequality, we geth≥0, with equality if only ify =z. From h(x) = 2

3S−x

3R^p−x^p 2

¹_p

−1 and

h⁰(x) = 3 (3S−x)²

3R^p−x^p 2

^1−p_p

(R^p−Sx^p−1)>0,

it follows that the functionh(x)is strictly increasing, andh(x)≥0impliesx≥x₁. In the caseh(0)≥ 0we havex₁ = 0, and in the caseh(0) <0we havex₁ >0and h(x₁) = 0.

Sub-casex = x₁. Ifh(0) ≥ 0, then0 = x₁ < y(x₁) ≤ z(x₁). If h(0) < 0, then h(x₁) = 0, and sinceh= 0impliesy=z, we have0< x₁ < y(x₁) = z(x₁).

Sub-case x > x₁. Since h(x) is strictly increasing, for x > x₁ we have h(x) >

h(x₁) ≥ 0, hence h(x) > 0and y < z. From x+y(x) +z(x) = 3S and x^p + y^p(x) +z^p(x) = 3R^p, we get

y⁰(x) = x^p−1−z^p−1

z^p−1−y^p−1, z⁰(x) = y^p−1−x^p−1 z^p−1−y^p−1.

Sincey⁰(x)<0, the functiony(x)is strictly decreasing. Taking account ofy(x₁)>

x₁ (see sub-casex = x₁), there existsx₂ > x₁ such thaty(x₂) = x₂, y(x) > xfor x₁ < x < x₂, andy(x) < xforx > x₂. The conditiony ≥ximpliesx₁ < x ≤x₂. We see now that z⁰(x) > 0 forx1 < x < x2. Consequently, the functionz(x) is

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strictly increasing, and hencez(x) > z(x₁) ≥ y(x₁) > y(x). Finally, we conclude thatx < y < zforx∈(x₁, x₂), andx=y < zforx=x₂.

Proof of Proposition1.2. Consider the function

F(x) =f(x) +f(y(x)) +f(z(x))

defined onx ∈ [x₁, x₂]. We claim thatF(x)is minimal forx =x₁ and is maximal forx=x₂. If this assertion is true, then by Lemma1.1it follows that:

(a) F(x)is minimal for0 < x = y < z in the casep ≤ 0, or for eitherx = 0or 0< x < y=zin the casep >1;

(b) F(x)is maximal for0< x=y < z.

In order to prove the claim, assume thatx ∈ (x₁, x₂). By Lemma 1.1, we have 0< x < y < z. From

x+y(x) +z(x) =a+b+c and x^p+y^p(x) +z^p(x) =a^p+b^p+c^p, we get

y⁰+z⁰ =−1, y^p−1y⁰+z^p−1z⁰ =−x^p−1, whence

y⁰ = x^p−1−z^p−1

z^p−1−y^p−1, z⁰ = x^p−1−y^p−1 y^p−1−z^p−1. It is easy to check that this result is also valid forp= 0. We have

F⁰(x) =f⁰(x) +y⁰f⁰(y) +z⁰f⁰(z)

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and

F⁰(x)

(x^p−1−y^p−1)(x^p−1−z^p−1)

= g(x^p−1)

(x^p−1−y^p−1)(x^p−1−z^p−1) + g(y^p−1)

(y^p−1−z^p−1)(y^p−1−x^p−1) + g(z^p−1)

(z^p−1−x^p−1)(z^p−1−y^p−1). Sinceg is strictly convex, the right hand side is positive. On the other hand,

(x^p−1−y^p−1)(x^p−1−z^p−1)>0.

These results imply F⁰(x) > 0. Consequently, the function F(x) is strictly increasing for x ∈ (x1, x2). Excepting the trivial case when p > 1, x1 = 0 and

u→0limf(u) =−∞, the functionF(x)is continuous on[x1, x2], and hence is minimal only forx=x₁, and is maximal only forx=x₂.

Proof of Theorem1.3. We will consider two cases.

Case p ∈ (−∞,0]∪(1,∞). Excepting the trivial case when p > 1, x₁ = 0 and

u→0limf(u) =−∞, the functionF_n(x₁, x₂, . . . , x_n)attains its minimum and maximum values, and the conclusion follows from Proposition 1.2 above, via contradiction.

For example, let us consider the casep ≤ 0. In order to prove thatFnis maximal for0 < x₁ =x₂ =· · ·=x_n−1 ≤x_n, we assume, for the sake of contradiction, that F_n attains its maximum at(b₁, b₂, . . . , b_n)withb₁ ≤ b₂ ≤ · · · ≤ b_n andb₁ < b_n−1. Letx₁, xn−1, x_nbe positive numbers such thatx₁+xn−1+x_n=b₁+bn−1+b_nand x^p₁+x^p_n−1+x^p_n=b^p₁+b^p_n−1+b^p_n. According to Proposition1.2, the expression

F3(x1, xn−1, xn) =f(x1) +f(xn−1) +f(xn)

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is maximal only for x₁ = x_n−1 < x_n, which contradicts the assumption that F_n attains its maximum at(b₁, b₂, . . . , b_n)withb₁ < bn−1.

Casep∈(0,1). This case reduces to the casep >1, replacing each of thea_ibya

1 p

i , each of thex_i byx

1 p

i , and then pby ¹_p. Thus, we obtain the sufficient condition that h(x) = xf⁰

x^1−p¹

to be strictly convex on(0,∞). We claim that this condition is equivalent to the condition thatg(x) = f⁰

x^p−1¹

to be strictly convex on (0,∞).

Actually, for our proof, it suffices to show that ifg(x)is strictly convex on(0,∞), thenh(x) is strictly convex on (0,∞). To show this, we see that g ¹_x

= ¹_xh(x).

Sinceg(x)is strictly convex on(0,∞), by Jensen’s inequality we have ug

1 x

+vg

1 y

>(u+v)g ^u

x +^v_y u+v

for anyx, y, u, v >0withx6=y. This inequality is equivalent to u

xh(x) + v

yh(y)>

u x +v

y

h u+v

u x + ^v_y

! .

Substitutingu=txandv = (1−t)y, wheret∈(0,1), reduces the inequality to th(x) + (1−t)h(y)> h(tx+ (1−t)y),

which shows us thath(x)is strictly convex on(0,∞).

Proof of Corollary1.7. We will apply Theorem1.3 to the functionf(u) = plnu.

We see that lim

u→0f(u) = −∞forp > 0, and f⁰(u) = p

u, g(x) =f⁰ x^p−1¹

=px^1−p¹ , g⁰⁰(x) = p²

(1−p)²x^2p−1^1−p.

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Sinceg⁰⁰(x) > 0forx > 0, the functiong(x)is strictly convex on(0,∞), and the conclusion follows by Theorem1.3.

Proof of Corollary1.8. We will apply Theorem1.3to the function f(u) =q(q−1)(q−p)u^q.

Forp > 0, it is easy to check that eitherf(u)is continuous at u = 0 (in the case q >0) or lim

u→0f(u) = −∞(in the caseq <0). We have f⁰(u) =q²(q−1)(q−p)u^q−1 and

g(x) = f⁰

x^p−1¹

=q²(q−1)(q−p)x

q−1 p−1, g⁰⁰(x) = q²(q−1)²(q−p)²

(p−1)² x

2p−1 1−p.

Sinceg⁰⁰(x) > 0forx > 0, the functiong(x)is strictly convex on(0,∞), and the conclusion follows by Theorem1.3.

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3. Applications

Proposition 3.1. Letx, y, zbe non-negative real numbers such thatx+y+z = 2.

Ifr₀ ≤r ≤3, wherer₀ = _{ln 3−ln 2}^{ln 2} ≈1.71, then

x^r(y+z) +y^r(z+x) +z^r(x+y)≤2.

Proof. Rewrite the inequality in the homogeneous form

x^r+1+y^r+1+z^r+1+ 2

x+y+z 2

r+1

≥(x+y+z)(x^r+y^r+z^r), and apply Corollary1.8(casep=randq=r+ 1):

If0≤x≤y≤z such that

x+y+z =constant and x^r+y^r+z^r =constant,

then the sumx^r+1+y^r+1+z^r+1is minimal when eitherx= 0or0< x≤y=z.

Casex= 0. The initial inequality becomes

yz(y^r−1+z^r−1)≤2,

wherey+z = 2. Since0< r−1≤2, by the Power Mean inequality we have y^r−1+z^r−1

2 ≤

y²+z² 2

^r−1₂ . Thus, it suffices to show that

yz

y²+z² 2

^r−1₂

≤1.

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Taking account of

y²+z²

2 = 2(y²+z²)

(y+z)² ≥1 and r−1 2 ≤1, we have

1−yz

y²+z² 2

^r−1₂

≥1−yz

y²+z² 2

= (y+z)⁴

16 − yz(y²+z²) 2

= (y−z)⁴ 16 ≥0.

Case 0 < x ≤ y = z. In the homogeneous inequality we may leave aside the constraintx+y+z = 2, and consider y = z = 1, 0 < x ≤ 1. The inequality reduces to

1 + x

2 r+1

−x^r−x−1≥0.

Since 1 + ^x₂r+1

is increasing and x^r is decreasing in respect to r, it suffices to considerr=r₀. Let

f(x) = 1 + x

2 r0+1

−x^r⁰ −x−1.

We have

f⁰(x) = r0+ 1 2

1 + x

2 r0

−r₀x^r⁰⁻¹−1, 1

r₀f⁰⁰(x) = r0+ 1 4

1 + x

2 r0

− r0−1 x^2−r⁰ .

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Sincef⁰⁰(x)is strictly increasing on(0,1],f⁰⁰(0₊) = −∞and 1

r₀f⁰⁰(1) = r0+ 1 4

3 2

r0

−r₀+ 1

= r₀+ 1

2 −r₀+ 1 = 3−r₀ 2 >0,

there exists x1 ∈ (0,1) such that f⁰⁰(x1) = 0, f⁰⁰(x) < 0 for x ∈ (0, x1), and f⁰⁰(x) > 0for x ∈ (x1,1]. Therefore, the function f⁰(x)is strictly decreasing for x∈[0, x₁], and strictly increasing forx∈[x₁,1]. Since

f⁰(0) = r₀−1

2 >0 and f⁰(1) = r₀+ 1 2

3 2

r0

−2

= 0,

there existsx₂ ∈(0, x₁)such thatf⁰(x₂) = 0,f⁰(x)>0forx∈[0, x₂), andf⁰(x)<

0forx ∈ (x2,1). Thus, the function f(x)is strictly increasing forx ∈ [0, x2], and strictly decreasing forx∈ [x2,1]. Sincef(0) =f(1) = 0, it follows that f(x)≥ 0 for0< x≤1, establishing the desired result.

For x ≤ y ≤ z, equality occurs when x = 0 and y = z = 1. Moreover, for r=r₀, equality holds again whenx=y=z = 1.

Proposition 3.2 ([12]). Letx, y, zbe non-negative real numbers such thatxy+yz+ zx= 3. If1< r≤2, then

x^r(y+z) +y^r(z+x) +z^r(x+y)≥6.

Proof. Rewrite the inequality in the homogeneous form

x^r(y+z) +y^r(z+x) +z^r(x+y)≥6

xy+yz +zx 3

^r+1₂ .

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For convenience, we may leave aside the constraintxy+yz +zx = 3. Using now the constraintx+y+z = 1, the inequality becomes

x^r(1−x) +y^r(1−y) +z^r(1−z)≥6

1−x²−y²−z² 6

^r+1₂ .

To prove it, we will apply Corollary 1.4 to the function f(u) = −u^r(1−u) for 0≤u≤1. We havef⁰(u) = −ru^r−1+ (r+ 1)u^r and

g(x) =f⁰(x) =−rx^r−1+ (r+ 1)x^r, g⁰⁰(x) =r(r−1)x^r−3[(r+ 1)x+ 2−r].

Sinceg⁰⁰(x)>0forx >0,g(x)is strictly convex on[0,∞). According to Corollary 1.4, if0≤x≤y≤zsuch thatx+y+z = 1andx²+y²+z² =constant, then the sumf(x) +f(y) +f(z)is maximal for0≤x=y≤z.

Thus, we have only to prove the original inequality in the casex = y ≤z. This means, to prove that0< x≤1≤yandx²+ 2xz = 3implies

x^r(x+z) +xz^r ≥3.

Letf(x) =x^r(x+z) +xz^r−3,withz = ^3−x_2x².

Differentiating the equationx² + 2xz = 3yieldsz⁰ = ^−(x+z)_x . Then, f⁰(x) = (r+ 1)x^r+rx^r−1z+z^r+ (x^r+rxz^r−1)z⁰

= (x^r−1−z^r−1)[rx+ (r−1)z]≤0.

The function f(x) is strictly decreasing on [0,1], and hencef(x) ≥ f(1) = 0 for 0< x≤1. Equality occurs if and only ifx=y=z = 1.

Proposition 3.3 ([5]). Ifx₁, x₂, . . . , x_nare positive real numbers such that x₁+x₂+· · ·+x_n = 1

x₁ + 1

x₂ +· · ·+ 1 x_n,

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then 1

1 + (n−1)x₁ + 1

1 + (n−1)x₂ +· · ·+ 1

1 + (n−1)x_n ≥1.

Proof. We have to consider two cases.

Casen = 2. The inequality is verified as equality.

Casen ≥3. Assume that0< x₁ ≤ x₂ ≤ · · · ≤x_n, and then apply Corollary1.5to the functionf(u) = _1+(n−1)u¹ foru >0. We havef⁰(u) = _[1+(n−1)u]⁻⁽ⁿ⁻¹⁾ 2 and

g(x) = f⁰ 1

√x

= −(n−1)x (√

x+n−1)², g⁰⁰(x) = 3(n−1)²

2√ x(√

x+n−1)⁴.

Sinceg⁰⁰(x) > 0, g(x) is strictly convex on(0,∞). According to Corollary1.5, if 0< x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x₁+x₂ +· · ·+x_n=constant and 1

x₁ + 1

x₂ +· · ·+ 1

x_n =constant,

then the sumf(x₁) +f(x₂) +· · ·+f(x_n)is minimal when0 < x₁ ≤ x₂ = x₃ =

· · ·=x_n.

Thus, we have to prove the inequality 1

1 + (n−1)x + n−1

1 + (n−1)y ≥1, under the constraints0< x≤1≤yand

x+ (n−1)y= 1

x +n−1 y .

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The last constraint is equivalent to

(n−1)(y−1) = y(1−x²) x(1 +y). Since

1

1 + (n−1)x + n−1

1 + (n−1)y −1

= 1

1 + (n−1)x − 1

n + n−1

1 + (n−1)y − n−1 n

= (n−1)(1−x)

n[1 + (n−1)x] − (n−1)²(y−1) n[1 + (n−1)y]

= (n−1)(1−x)

n[1 + (n−1)x] − (n−1)y(1−x²) nx(1 +y)[1 + (n−1)y], we must show that

x(1 +y)[1 + (n−1)y]≥y(1 +x)[1 + (n−1)x], which reduces to

(y−x)[(n−1)xy−1]≥0.

Sincey−x≥0, we have still to prove that (n−1)xy≥1.

Indeed, fromx+ (n−1)y = ¹_x +ⁿ⁻¹_y we getxy= ^y+(n−1)x_x+(n−1)y, and hence (n−1)xy−1 = n(n−2)x

x+ (n−1)y >0.

Forn≥3, one has equality if and only ifx1 =x2 =· · ·=xn= 1.

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Proposition 3.4 ([10]). Leta₁, a₂, . . . , a_nbe positive real numbers such thata₁a₂· · ·a_n = 1. Ifmis a positive integer satisfyingm≥n−1, then

a^m₁ +a^m₂ +· · ·+a^m_n + (m−1)n≥m 1

a₁ + 1

a₂ +· · ·+ 1 a_n

. Proof. Forn= 2(hencem ≥1), the inequality reduces to

a^m₁ +a^m₂ + 2m−2≥m(a₁+a₂).

We can prove it by summing the inequalitiesa^m₁ ≥ 1 +m(a₁ −1)anda^m₂ ≥ 1 + m(a₂ −1), which are straightforward consequences of Bernoulli’s inequality. For n≥3, replacinga1, a2, . . . , anby _x¹

1,_x¹

2, . . . ,_x¹

n, respectively, we have to show that 1

x^m₁ + 1

x^m₂ +· · ·+ 1

x^m_n + (m−1)n ≥m(x1+x2 +· · ·+xn)

forx₁x₂· · ·x_n= 1. Assume0< x₁ ≤x₂ ≤ · · · ≤x_nand apply Corollary1.8(case p= 0andq=−m):

If0< x₁ ≤x₂ ≤ · · · ≤x_nsuch that

x1+x2+· · ·+xn =constant and x₁x₂· · ·x_n = 1,

then the sum_x¹m 1 +_x¹m

2 +· · ·+_x¹m

n is minimal when0< x1 =x2 =· · ·=xn−1 ≤xn. Thus, it suffices to prove the inequality for x₁ = x₂ = · · · = xn−1 = x ≤ 1, x_n=yandxⁿ⁻¹y= 1, when it reduces to:

n−1 x^m + 1

y^m + (m−1)n≥m(n−1)x+my.

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By the AM-GM inequality, we have n−1

x^m + (m−n+ 1) ≥ m

xⁿ⁻¹ =my.

Then, we have still to show that 1

y^m −1≥m(n−1)(x−1).

This inequality is equivalent to

x^mn−m−1−m(n−1)(x−1)≥0 and

(x−1)[(x^mn−m−1−1) + (x^mn−m−2−1) +· · ·+ (x−1)] ≥0.

The last inequality is clearly true. Forn = 2 andm = 1, the inequality becomes equality. Otherwise, equality occurs if and only ifa₁ =a₂ =· · ·=a_n= 1.

Proposition 3.5 ([6]). Let x₁, x₂, . . . , x_n be non-negative real numbers such that x₁ +x₂ +· · ·+x_n = n. If k is a positive integer satisfying 2 ≤ k ≤ n+ 2, and r= _n−1ⁿ k−1

−1, then

x^k₁ +x^k₂+· · ·+x^k_n−n≥nr(1−x₁x₂· · ·x_n).

Proof. Ifn = 2, then the inequality reduces tox^k₁ +x^k₂ −2 ≥ (2^k−2)x₁x₂. For k = 2and k = 3, this inequality becomes equality, while fork = 4 it reduces to 6x₁x₂(1−x₁x₂)≥0, which is clearly true.

Consider now n ≥ 3 and 0 ≤ x₁ ≤ x₂ ≤ · · · ≤ x_n. Towards proving the inequality, we will apply Corollary1.7(casep=k >0): If0≤x₁ ≤x₂ ≤ · · · ≤x_n

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such thatx₁+x₂+· · ·+x_n=nandx^k₁+x^k₂+· · ·+x^k_n=constant, then the product x₁x₂· · ·x_nis minimal when eitherx₁ = 0or0< x₁ ≤x₂ =x₃ =· · ·=x_n.

Casex₁ = 0. The inequality reduces to

x^k₂+· · ·+x^k_n ≥ n^k (n−1)^k−1,

withx₂+· · ·+x_n = n, This inequality follows by applying Jensen’s inequality to the convex functionf(u) =u^k:

x^k₂+· · ·+x^k_n ≥(n−1)

x₂+· · ·+x_n n−1

k

.

Case0 < x₁ ≤x₂ =x₃ =· · · =x_n. Denotingx₁ =xandx₂ =x₃ =· · · =x_n = y, we have to prove that for0 < x ≤ 1 ≤ y andx+ (n−1)y = n, the inequality holds:

x^k+ (n−1)y^k+nrxyⁿ⁻¹−n(r+ 1) ≥0.

Write the inequality asf(x)≥0, where

f(x) =x^k+ (n−1)y^k+nrxyⁿ⁻¹−n(r+ 1), with y= n−x n−1. We see thatf(0) =f(1) = 0. Sincey⁰ = _n−1⁻¹ , we have

f⁰(x) =k(x^k−1−y^k−1) +nryⁿ⁻²(y−x)

= (y−x)[nryⁿ⁻²−k(y^k−2+y^k−3x+· · ·+x^k−2)]

= (y−x)yⁿ⁻²[nr−kg(x)], where

g(x) = 1

y^n−k + x

y^n−k+1 +· · ·+x^k−2 yⁿ⁻².

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Since the function y(x) = ^n−x_n−1 is strictly decreasing, the function g(x) is strictly increasing for2≤k ≤n. Fork =n+ 1, we have

g(x) =y+x+ x²

y +· · ·+xⁿ⁻¹ yⁿ⁻²

= (n−2)x+n n−1 +x²

y +· · ·+xⁿ⁻¹ yⁿ⁻², and fork=n+ 2, we have

g(x) =y²+yx+x²+x³

y +· · ·+ xⁿ yⁿ⁻²

= (n²−3n+ 3)x²+n(n−3)x+n² (n−1)² +x³

y +· · ·+ xⁿ yⁿ⁻².

Therefore, the functiong(x)is strictly increasing for2≤k ≤n+2, and the function h(x) =nr−kg(x)

is strictly decreasing. Note that

f⁰(x) = (y−x)yⁿ⁻²h(x).

We assert thath(0) > 0and h(1) < 0. If our claim is true, then there exists x₁ ∈ (0,1)such thath(x₁) = 0, h(x)> 0forx ∈ [0, x₁), andh(x)< 0forx ∈ (x₁,1].

Consequently,f(x)is strictly increasing forx ∈ [0, x₁], and strictly decreasing for x∈[x₁,1]. Sincef(0) =f(1) = 0, it follows thatf(x)≥ 0for0< x≤1, and the proof is completed.

In order to prove that h(0) > 0, we assume thath(0) ≤ 0. Then,h(x) < 0for x ∈ (0,1), f⁰(x) < 0 forx ∈ (0,1), andf(x)is strictly decreasing for x ∈ [0,1],

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which contradicts f(0) = f(1). Also, ifh(1) ≥ 0, then h(x) > 0for x ∈ (0,1), f⁰(x) > 0forx ∈ (0,1), and f(x) is strictly increasing forx ∈ [0,1], which also contradictsf(0) =f(1).

Forn≥3andx₁ ≤x₂ ≤ · · · ≤x_n, equality occurs whenx₁ =x₂ =· · ·=x_n= 1, and also whenx₁ = 0andx₂ =· · ·=x_n = _n−1ⁿ .

Remark 2. Fork = 2,k= 3andk = 4, we get the following nice inequalities:

(n−1)(x²₁ +x²₂+· · ·+x²_n) +nx₁x₂· · ·x_n≥n², (n−1)²(x³₁+x³₂+· · ·+x³_n) +n(2n−1)x₁x₂· · ·x_n ≥n³, (n−1)³(x⁴₁+x⁴₂+· · ·+x⁴_n) +n(3n²−3n+ 1)x₁x₂· · ·x_n ≥n⁴.

Remark 3. The inequality for k = n was posted in 2004 on the Mathlinks Site - Inequalities Forum by Gabriel Dospinescu and C˘alin Popa.

Proposition 3.6 ([11]). Let x₁, x₂, . . . , x_nbe positive real numbers such that _x¹

1 +

1

x2 +· · ·+ _x¹

n =n. Then

x₁+x₂+· · ·+x_n−n ≤en−1(x₁x₂· · ·x_n−1), wheree_n−1 = 1 + _n−1¹ ⁿ⁻¹

< e.

Proof. Replacing each of thex_iby _a¹

i, the statement becomes as follows:

Ifa₁, a₂, . . . , a_nare positive numbers such thata₁+a₂+· · ·+a_n =n, then a1a2· · ·an

1 a₁ + 1

a₂ +· · ·+ 1

a_n −n+en−1

≤en−1.

It is easy to check that the inequality holds forn = 2. Consider nown ≥3, assume that0< a1 ≤ a2 ≤ · · · ≤ an and apply Corollary1.7 (casep= −1): If0 < a1 ≤

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a₂ ≤ · · · ≤a_nsuch thata₁+a₂+· · ·+a_n =nand _a¹

1 +_a¹

2 +· · ·+_a¹

n =constant, then the producta₁a₂· · ·a_nis maximal when0< a₁ ≤a₂ =a₃ =· · ·=a_n.

Denoting a₁ = x and a₂ = a₃ = · · · = a_n = y, we have to prove that for 0< x≤1≤y < _n−1ⁿ andx+ (n−1)y =n, the inequality holds:

yⁿ⁻¹+ (n−1)xyⁿ⁻²−(n−en−1)xyⁿ⁻¹ ≤en−1. Letting

f(x) =yⁿ⁻¹+ (n−1)xyⁿ⁻²−(n−en−1)xyⁿ⁻¹−en−1, with y= n−x

n−1,

we must show thatf(x) ≤ 0for0 < x ≤ 1. We see thatf(0) = f(1) = 0. Since y⁰ = _n−1⁻¹ , we have

f⁰(x)

yⁿ⁻³ = (y−x)[n−2−(n−en−1)y] = (y−x)h(x), where

h(x) =n−2−(n−en−1)n−x n−1 is a linear increasing function.

Let us show that h(0) < 0 and h(1) > 0. If h(0) ≥ 0, then h(x) > 0 for x ∈ (0,1), hence f⁰(x) > 0 for x ∈ (0,1), and f(x) is strictly increasing for x∈[0,1], which contradictsf(0) =f(1). Also,h(1) =en−1−2>0.

From h(0) < 0and h(1) > 0, it follows that there existsx₁ ∈ (0,1)such that h(x₁) = 0, h(x) < 0forx ∈ [0, x₁), andh(x) > 0for x ∈ (x₁,1]. Consequently, f(x) is strictly decreasing for x ∈ [0, x₁], and strictly increasing for x ∈ [x₁,1].

Sincef(0) =f(1) = 0, it follows thatf(x)≤0for0≤x≤1.

Forn ≥3, equality occurs whenx1 =x2 =· · ·=xn = 1.