New inequalities for the triangle

Vol. 17, No.1, April 2009, pp 70-89 ISSN 1222-5657, ISBN 978-973-88255-5-0, www.hetfalu.ro/octogon

70

New inequalities for the triangle

Mih´aly Bencze, Nicu¸sor Minculete and Ovidiu T. Pop⁵

ABSTRACT.In this paper we will prove some new inequalities for the triangle.

Among these, we will improve Euler’s Inequality, Mitrinovi´c’s Inequality and Weitzenb¨ock’s Inequality, thus:

R≥ 4

P

cyclic

r Fλ

1 ha,_h¹_b

(n)

≥2r; s≥1 2

X

cyclic

pFλ(s−a, s−b) (n)≥3√ 3r;

and

a^2α+b^2α+c^2α≥1 2

X

cyclic

pFλ(a^2α, b^2α) (n)≥3 4∆

√3 α

,

whereFλ(x, y) (n) = [(1 + (1−2λ)ⁿ)x+ (1−(1−2λ)ⁿ)y]·

·[(1−(1−2λ)ⁿ) + (1 + (1−2λ)ⁿ)y],withλ∈[0,1],for anyx, y≥0 and for all integersn≥0.

1. INTRODUCTION

Among well known the geometric inequalities, we recall the famous inequality of Euler,R≥2r, the inquality of Mitrinovi´c,s≥3√

3r, and in the year 1919 Weitzenb¨ock published in Mathematische Zeitschrift the following inequality,

a²+b²+c² ≥4√ 3∆.

This inequality later, in 1961, was given at the International Mathematical Olympiad. In 1927, this inequality appeared as the generalization

∆≤

√3 4

a^k+b^k+c^k 3

²_k ,

in one of the issues of theAmerican Mathematical Monthly. Fork= 2,we obtain the Weitzenb¨ock Inequality.

In this paper we will prove several improvements for these inqualities.

5Received: 17.03.2009

2000Mathematics Subject Classification. 26D05, 26D15, 51M04

Key words and phrases. Geometric inequalities, Euler‘s inequality, Mitrinovic‘s inequality, Weitzenb¨ok inequality.

2. MAIN RESULTS

In the following, we will use the notations: a, b, c−the lengths of the sides, ha, hb, hc−the lengths of the altitudes,ra, rb, rc−the radii of the excircles, s is the semi-perimeter; R is the circumradius,r−the inradius, and ∆− the area of the triangleABC.

Lemma 2.1 Ifx, y≥0 andλ∈[0,1], then the inequality x+y

2 2

≥[(1−λ)x+λy]·[λx+ (1−λ)y]≥xy (2.1) holds.

Proof. The inequality x+y

2 2

≥[(1−λ)x+λy]·[λx+ (1−λ)y]

is equivalent to

(1−2λ)²x²−2 (1−2λ)²xy+ (1−2λ)²y² ≥0, which means that

(1−2λ)²(x−y)² ≥0,

which is true. The equality holds if and only if λ= ¹₂ orx=y.

The inequality

[(1−λ)x+λy]·[λx+ (1−λ)y]≥xy becomes

λ(1−λ)x²−2λ(1−λ)xy+λ(1−λ)y²≥0 Therefore, we obtain

λ(1−λ) (x−y)²≥0,

which is true, becauseλ∈[0,1]. The equality holds if and only if λ∈ {0,1} orx=y.

We consider the expression

F_λ(x, y) (n) = [(1 + (1−2λ)ⁿ)x+ (1−(1−2λ)ⁿ)y]·

·[(1−(1−2λ)ⁿ)x+ (1 + (1−2λ)ⁿ)y], withλ∈[0,1],for anyx, y≥0, and for all integers n≥0.

Theorem 2.2 There are the following relations:

F_λ((1−λ)x+λy, λx+ (1−λ)y) (n) =F_λ(x, y) (n+ 1) ; (2.2) F_λ(x, y) (n+ 1)≥F_λ(x, y) (n) (2.3) and

(x+y)²≥F_λ(x, y) (n)≥4xy, (2.4) for anyλ∈[0,1] ,for anyx, y≥0 and all integersn≥0.

Proof. We make the following calculation:

F_λ((1−λ)x+λy, λx+ (1−λ)y) (n) =

= [(1 + (1−2λ)ⁿ) ((1−λ)x+λy) + (1−(1−2λ)ⁿ) (λx+ (1−λ)y)]·

·[(1−(1−2λ)ⁿ) ((1−λ)x+λy) + (1 + (1−2λ)ⁿ) (λx+ (1−λ)y)] =

={[1−λ+ (1−λ) (1−2λ)ⁿ+λ−λ(1−2λ)ⁿ]x+

+ [λ+λ(1−2λ)ⁿ+ 1−λ−(1−λ) (1−2λ)ⁿ]y}

· {[1−λ−(1−λ) (1−2λ)ⁿ+λ+λ(1−2λ)ⁿ]x+

+ [λ−(1−2λ)ⁿ+ 1−λ+ (1−λ) (1−2λ)ⁿ]y}=

=h

1 + (1−2λ)ⁿ⁺¹ x+

1−(1−2λ)ⁿ⁺¹ yi h

1−(1−2λ)ⁿ⁺¹ x+

1 + (1−2λ)ⁿ⁺¹ yi

=

=F_λ(x, y) (n+ 1),

soF_λ((1−λ)x+λy, λx+ (1−λ)y) (n) =F_λ(x, y) (n+ 1). We use the induction onn. Forn= 0,we obtain the inequality

F_λ(x, y) (1)≥F_λ(x, y) (0). Therefore, we deduce the following inequality:

4 [(1−λ)x+λy]·[λx+ (1−λ)y]≥4xy, which is true, from Lemma 2.1.

We assume it is true for every integer ≤n, so

F_λ(x, y) (n+ 1)≥F_λ(x, y) (n) We will prove that

F_λ(x, y) (n+ 2)≥F_λ(x, y) (n+ 1). (2.5) Using the substitutions x→(1−λ)x+λy andy →λx+ (1−λ)y in the inequality (2.3), we deduce

F_λ((1−λ)x+λy, λx+ (1−λ)y) (n+ 1)≥

≥F_λ((1−λ)x+λy, λx+ (1−λ)y) (n), so, from equality (2.2), we have

F_λ(x, y) (n+ 2)≥F_λ(x, y) (n+ 1). so we obtain (2.6).

According to inequality (2.3), we can write the sequence of inequalities

F_λ(x, y) (n)≥F_λ(x, y) (n−1)≥...≥F_λ(x, y) (1)≥F_λ(x, y) (0) = 4xy.

Therefore, we have

F_λ(x, y) (n)≥4xy, for anyλ∈[0,1], x, y≥0 and for all integersn≥0.

Ifλ∈(0,1), then 1−2λ∈(−1,1) and passing to limit whenn→ ∞, we obtain

nlim→∞F_λ(x, y) (n) = (x+y)².

Since the sequence (F_λ(x, y) (n))_n≥0 is increasing, we deduce

(x+y)²≥F_λ(x, y) (n), for anyλ∈(0,1), x, y≥0 and for all integersn≥0.

From the inequalities above we have that

(x+y)²≥F_λ(x, y) (n)≥4xy, for anyλ∈(0,1), x, y≥0 and for all integers n≥0.

Ifλ= 0 and λ= 1, then F_λ(x, y) (n) = 4xy, so (x+y)²≥F_λ(x, y) (n)≥4xy.

It follows that

(x+y)²≥F_λ(x, y) (n)≥4xy, for anyλ∈[0,1], x, y≥0 and for all integers n≥0.

Thus, the proof of Theorem 2.2 is complete.

Remark 1. It is easy to see that there is the sequence of inequalities

(x+y)² ≥...≥F_λ(x, y) (n)≥F_λ(x, y) (n−1)≥...

...≥F_λ(x, y) (1)≥F_λ(x, y) (0) = 4xy. (2.6) Corollary 2.3. There are the following inequalities:

x+y≥p

F_λ(x, y) (n)≥2√

xy; (2.7)

x²+y²≥p

F_λ(x², y²) (n)≥2xy; (2.8) x+y+z≥ 1

2 X

cyclic

pF_λ(x, y) (n)≥√xy+√yz+√

zx; (2.9)

x²+y²+z²≥ 1 2

X

cyclic

pF_λ(x², y²) (n)≥xy+yz+zx; (2.10)

x²+y²+z²+xy+yz+zx≥ 1 2

X

cyclic

F_λ(x, y) (n)≥2 (xy+yz+zx) (2.11) and

(x+y) (y+z) (z+x)≥s Y

cyclic

F_λ(x, y) (n)≥8xyz, (2.12) for anyλ∈[0,1],for anyx, y≥0, and for all integers n≥0.

Proof. From Theorem 2.2, we easily deduce inequality (2.7). Using the substitutions x→x² and y→y² in inequality (2.7), we obtain inequality (2.8). Similarly to inequality (2.7), x+y≥p

F_λ(x, y) (n)≥2√xy, we can write the following inequalities:

y+z≥p

F_λ(y, z) (n)≥2√yz and z+x≥p

F_λ(z, x) (n)≥2√ zx, which means, by adding, that

x+y+z≥ 1 2

X

cyclic

pF_λ(x, y) (n)≥√xy+√yz+√ zx.

It is easy to see that, by making the substitutions x→x² andy →y² in inequality (2.9), we obtain inequality (2.10). Similar to inequality (2.4), (x+y)²≥F_λ(x, y) (n)≥4xy, we obtain the following inequalities:

(y+z)²≥F_λ(y, z) (n)≥4yz and (z+x)² ≥F_λ(z, x) (n)≥4zx.

By adding them, we have inequality (2.11) and by multiplying them, we obtain inequality (2.12).

Lemma 2.4 For any triangle ABC, the following inequality,

√ab+√ bc+√

ca≥ 4∆

R , (2.13)

holds.

Proof. We apply the arithmetic-geometric mean inequality and we find that

√ab+√ bc+√

ca≥3√³ abc.

It is sufficient to show that

3√³

abc≥ 4∆

R . (2.14)

Inequality (2.14) is equivalent to

27abc≥ 64∆³ R³ , so

27·4R∆≥ 64∆³ R³ , which means that

27R⁴≥16∆². (2.15)

Using Mitrinovi´c’s Inequality, 3√

3R≥2s, and Euler’s InequalityR≥2r, we deduce, by multiplication, that

3√

3R² ≥4∆.

It follows (2.15).

Corollary 2.5. In any triangle ABC, there are the following inequalities:

R≥ 4

P

cyclic

r F_λ

1 ha,_h¹

b

(n)

≥2r; (2.16)

s≥ 1 2

X

cyclic

pF_λ(s−a, s−b) (n)≥3√

3r (2.17)

and

a^2α+b^2α+c^2α≥ 1 2

X

cyclic

pF_λ(a^2α, b^2α) (n)≥3 4∆

√3 α

, (2.18)

for anyλ∈[0,1], x, y≥0,n≥0 and α is a real numbers.

Proof. Making the substitutions x= _h¹

a, y= _h¹

b and z= _h¹

c in inequality (2.9), we obtain

1 h_a+ 1

h_b+ 1 h_c ≥ 1

2 X

cyclic

s F_λ

1 h_a, 1

h_b

(n)≥ 1

√h_ah_b+ 1

√h_bh_c+ 1

√h_ch_a. (2.19)

According to the equalities h_a= 2∆

a , h_b= 2∆

b andh_c = 2∆

c , we have

√1 hahb

+ 1

√hbhc

+ 1

√hcha

= 1 2∆

√ ab+√

bc+√ ca

. From Lemma 2.4, we deduce

√1

h_ah_b + 1

√h_bh_c + 1

√h_ch_a ≥ 2 R. If we use the identity

1 ha

+ 1 h_b + 1

hc

= 1 r

and inequality from above then inequality (2.18) becomes 1

r ≥ 1 2

X

cyclic

s F_λ

1 h_a, 1

h_b

(n)≥ 2

R. (2.20)

Consequently the inequalities (2.16) follows.

If in inequality (2.9) we takex=s−a, y=s−band z=s−c, then we deduce the inequality

s≥ 1 2

X

cyclic

pF_λ(s−a, s−b) (n)≥

≥p

(s−a) (s−b) +p

(s−b) (s−c) +p

(s−c) (s−a). (2.21) But, we know the identity P

cyclic

p(s−a, s−b) = P

cyclic

√bcsin^A₂. Using the arithmetic-geometric mean inequality, we obtain

X

cyclic

√bcsinA

2 ≥ 3³ r

abcsinA 2 sinB

2 sinC 2 = 3³

r

4R∆· r 4R =

= 3√³

∆r = 3√³

sr²≥3³ q

3√

3r³ = 3√ 3r.

Hence,

p(s−a) (s−b) +p

(s−b) (s−c) +p

(s−c) (s−a)≥3√

3r, (2.22) which means, according to inequalities (2.21) and (2.22), that

s≥ 1 2

X

cyclic

pF_λ(s−a, s−b) (n)≥3√ 3r.

Making the substitutions x=a^α, y=b^α, and z=c^α in inequality (2.9), we obtain the following inequality:

a^2α+b^2α+c^2α ≥ 1 2

X

cyclic

pF_λ(a^2α, b^2α) (n)≥a^αb^α+b^αc^α+c^αa^α. (2.23)

Applying the arithmetic- geometric mean inequality and P´olya-Szeg˝o’s Inequality, √³

a²b²c² ≥ ^4∆√₃, we deduce a^αb^α+b^αc^α+c^αa^α ≥ ³

q

(a²b²c²)^α = 3√₃

a²b²c²α

≥3 4∆

√3 α

,

so

a^αb^α+b^αc^α+c^αa^α≥3 4∆

√3 α

. (2.24)

According to inequalities (2.23) and (2.24), we obtain the inequality a^2α+b^2α+c^2α≥ 1

2 X

cyclic

pF_λ(a^2α, b^2α) (n)≥3 4∆

√3 α

.

Thus, the statement is true.

Remark 2. a) Inequality (2.16) implies the sequence of inequalities

R≥ 4

P

cyclic

r F_λ

1 ha,_h¹

b

(0)

≥...≥ 4 P

cyclic

r F_λ

1 ha,_h¹

b

(n−1)

≥

≥ 4

P

cyclic

r F_λ

1 ha,_h¹

b

(n)

≥...≥2r (2.25)

b) For α= 1 in inequality (2.18), we obtain

a²+b²+c²≥ 1 2

X

cyclic

pF_λ(a², b²) (n)≥4√

3∆, (2.26)

which proves Weitzenb¨ock’s Inequality, namely a²+b²+c² ≥4√

3∆.

Corollary 2.6. For any triangle ABC, there are the following inequalities:

2 s²−r²−4Rr

≥ 1 2

X

cyclic

pF_λ(a², b²) (n)≥s²+r²+ 4Rr, (2.27)

s²−2r²−8Rr≥ 1 2

X

cyclic

r F_λ

(s−a)²,(s−b)²

(n)≥r(4R+r), (2.28)

s²+r²+ 4Rr2

−8s²Rr

4R² ≥ 1

2 X

cyclic

q

F_λ h²_a, h²_b

(n)≥ 2s²r

R , (2.29) (4R+r)²−2s²≥ 1

2 X

cyclic

q

F_λ r²_a, r²_b

(n)≥s², (2.30)

8R²+r²−s²

8R² ≥ 1

2 X

cyclic

s F_λ

sin⁴ A

2,sin⁴ B 2

(n)≥ s²+r²−8Rr

16R² (2.31) and

(4R+r)²−s²

4R² ≥ 1

2 X

cyclic

s F_λ

cos⁴A

2,cos⁴B 2

(n)≥

≥ s²+ (4R+r)²

8R² . (2.32)

Proof. According to Corollary 2.3 we have the inequality x²+y²+z²≥ 1

2 X

cyclic

pF_λ(x², y²) (n)≥xy+yz+zx.

Using the substitutions

(x, y, z) ∈ {(a, b, c),(s−a, s−b, s−c),(ha, h_b, hc),(ra, r_b, rc),

sin²A

2,sin²B

2,sin²C 2

,

cos² A

2,cos²B

2,cos²C 2

,

we deduce the inequalities required.

Corollary 2.7. In any triangle ABC, there are the following inequalities:

3s²−r²−4Rr≥ 1 2

X

cyclic

F_λ(a, b) (n)≥2 s²+r²+ 4Rr

, (2.33)

s²−r²−4Rr≥ 1 2

X

cyclic

F_λ(s−a, s−b) (n)≥2r(4R+r), (2.34) s²+r²+ 4Rr2

−8s²Rr

4R² ≥ 1

2 X

cyclic

F_λ(h_a, h_b) (n)≥ 4s²r

R (2.35)

and

(4R+r)²−s²≥ 1 2

X

cyclic

F_λ(ra, r_b) (n)≥2s² (2.36) Proof. According to Corollary 2.3, we have the inequality

x²+y²+z²+xy+yz+zx≥ 1 2

X

cyclic

F_λ(x, y) (n)≥2 (xy+yz+zx). Using the substitutions

(x, y, z)∈ {(a, b, c),(s−a, s−b, s−c),(h_a, h_b, h_c),(r_a, r_b, r_c)} we deduce the inequalities from the statement.

Corollary 2.8. For any triangle ABC there are the following inequalities:

2s s²+r²+ 2Rr

≥ Y

cuclic

pF_λ(a, b) (n)≥32sRr, (2.37)

4sRr≥ Y

cyclic

pF_λ((s−a),(s−b)) (n)≥8sr², (2.38)

s²r s²+r²+ 4Rr

R² ≥ Y

cyclic

pF_λ(h_a, h_b) (n)≥ 16s²r²

R , (2.39)

4s²R≥ Y

cyclic

pF_λ(r_a, r_b) (n)≥8s²r, (2.40)

(2R−r) s²+r²−8Rr

−2Rr²

32R³ ≥ Y

cyclic

s F_λ

sin² A

2,sin²B 2

(n)≥

≥ r²

2R² (2.41)

and

(4R+r)³+s²(2R+r)

32R³ ≥ Y

cyclic

s Fλ

cos²A

2,cos²B 2

(n)≥ s²

2R² (2.42) Proof. According to Corollary 2.3, we have the inequality

(x+y) (y+z) (z+x)≥s Y

cyclic

F_λ(x, y) (n)≥8xyz.

Using the substitutions

(x, y, z) ∈ {(a, b, c),(s−a, s−b, s−c),(ha, hb, hc),(ra, rb, rc),

sin²A

2,sin²B

2,sin²C 2

,

cos² A

2,cos²B

2,cos²C 2

,

we deduce the inequalities required.

Remark 3. From Corollary 2.7, we obtain the inequality

2s s²+r²+ 2Rr

≥ Y

cyclic

pF_λ(a, b) (n)≥32sRr≥

≥ 8 Y

cyclic

pF_λ((s−a),(s−b)) (n)≥64sr².

We consider the expression

G(x, y) (n) = xy xⁿ⁻¹+yⁿ⁻¹

xⁿ⁺¹+yⁿ⁺¹

(xⁿ+yⁿ)² , (2.43)

where x, y >0 and for all integers n≥0.

Theorem 2.9. For anyx, y >0 and for all integers n≥0, there are the following relations:

a)

x+y 2

2

≥G(x, y) (n)≥xy (2.44) and

b)G(x, y) (n+ 1)≤G(x, y) (n). (2.45) Proof. We take λ= _xn^x+y^{n n},for all integersn≥0, in inequality (2.1), because λ∈(0,1), and we deduce

x+y 2

2

≥ xy xⁿ⁻¹+yⁿ⁻¹

xⁿ⁺¹+yⁿ⁺¹ (xⁿ+yⁿ)² ≥xy, so,

x+y 2

2

≥G(x, y) (n)≥xy.

To prove inequality (2.45), we can write G(x, y) (n+ 1)

G(x, y) (n) −1 =

=−(xy)ⁿ⁻¹(x−y)² x²+xy+y²

x²ⁿ+x²ⁿ⁻¹y+x²ⁿy²+...+y²ⁿ (xⁿ⁺¹+yⁿ⁺¹)³(xⁿ⁻¹+yⁿ⁻¹) ≤0.

Consequently, we have

G(x, y) (n+ 1)≤G(x, y) (n).

Remark 4. It is easy to see that there is the sequence of inequalities (x+y)²

4 =G(x, y) (0)≥G(x, y) (1)≥...

≥G(x, y) (n−1)≥G(x, y) (n)≥...≥xy. (2.46) Corollary 2.10

There are the following inequalities:

x+y

2 ≥p

G(x, y) (n)≥√xy; (2.47) x²+y²

2 ≥p

G(x², y²) (n)≥√

xy; (2.48)

x+y+z≥ X

cyclic

pG(x, y) (n)≥√

xy+√ yz+√

zx; (2.49) x²+y²+z² ≥ X

cyclic

pG(x², y²) (n)≥xy+yz+zx; (2.50)

1

2 x²+y²+z²+xy+yz+zx

≥ X

cyclic

G(x, y) (n)≥xy+yz+zx (2.51) and

1

8(x+y) (y+z) (z+x)≥s Y

cyclic

G(x, y) (n)≥xyz, (2.52) for anyx, y >0 and for all integers n≥0.

Proof. From Theorem 2.9, we easily deduce inequality (2.47). Using the substitutions x→x² and y→y² in inequality (2.47), we obtain inequality (2.48). Similarly to inequality (2.47), ^x+y₂ ≥p

G(x, y) (n)≥√xy, we can write the following inequalities:

y+z

2 ≥p

G(y, z) (n)≥√

yz and z+x

2 ≥p

G(z, x) (n)≥√ zx, which means, by adding, that

x+y+z≥ X

cyclic

pG(x, y) (n)≥√xy+√yz+√ zx.

It is easy to see that by making the substitutionsx→x² and y→y² in inequality (2.49), we obtain inequality (2.50). Similarly to inequality (2.44),

x+y 2

2

≥G(x, y) (n)≥xy, we obtain the following inequalities:

y+z 2

2

≥G(y, z) (n)≥yz and

z+x 2

2

≥G(z, x) (n)≥zx.

By adding them, we have inequality (2.51) and by multiplying them, we obtain inequality (2.52).

Corollary 2.11. In any triangleABC , there are the following inequalities:

R≥ 2

P

cyclic

r G

1 ha,_h¹

b

(n)

≥2r; (2.53)

s≥ X

cyclic

pG(s−a, s−b) (n)≥3√

3r (2.54)

and

a^2α+b^2α+c^2α≥ X

cyclic

pG(a^2α, b^2α) (n)≥3 4∆

√3 α

, (2.55)

for anyn≥0 and for every real numbersα.

Proof. Making the substitutions x= _h¹

a, y= _h¹

b and z= _h¹

c in inequality (2.49), we obtain

1 h_a + 1

h_b + 1

h_c ≥ X

cyclic

s G

1 h_a, 1

h_b

(n)≥ 1

√hah_b + 1

√h_bhc

+ 1

√h_ch_a. (2.56) From inequality (2.13), we have

√1

h_ah_b + 1

√h_bh_c + 1

√h_ch_a ≥ 2 R, and from the identity

1 h_a + 1

h_b + 1 h_c = 1

r we deduce

1

r ≥ X

cyclic

s G

1 h_a, 1

h_b

(n)≥ 2

R. (2.57)

Consequently

R≥ 2

P

cyclic

r G

1 ha,_h¹

b

(n)

≥2r.

If in inequality (2.49) we takex=s−a, y=s−b andz=s−c, then we deduce the inequality

s≥ X

cyclic

pG(s−a, s−b) (n)≥p

(s−a) (s−b)+

+p

(s−b) (s−c) +p

(s−c) (s−a). (2.58) But

p(s−a) (s−b) +p

(s−b) (s−c) +p

(s−c) (s−a)≥3√ 3r, which means that

s≥ X

cyclic

pG(s−a, s−b) (n)≥3√ 3r.

Making the substitutions , and in inequality (2.49), we obtain the following inequality:

a^2α+b^2α+c^2α ≥ X

cyclic

pG(a^2α, b^2α) (n)≥a^αb^α+b^αc^α+c^αa^α. (2.59)

Applying the arithmetic-geometric mean inequality and P´olya-Szeg¨o’s Inequality, √³

a²b²c² ≥ ^4∆√₃, we deduce

a^αb^α+b^αc^α+c^αa^α ≥ 4∆

√3 α

.

Therefore

s≥ X

cyclic

pG(s−a, s−b) (n)≥3√ 3r.

Remark 5. a) Inequality (2.53) implies the sequence of inequalities

R≥... 2

P

cyclic

r G

1 ha,_h¹

b

(n)

≥ 2

P

cyclic

r G

1 ha,_h¹

b

(n−1)

≥...

≥ 2 P

cyclic

r1 ha,_h¹

b

(0)

≥2r (2.60)

b) For α= 1 in inequality (2.55), we obtain a²+b²+c²≥ X

cyclic

pG(a², b²) (n)≥4√

3∆, (2.61)

which proves Weitzenb¨ock’s Inequality, namely a²+b²+c² ≥4√

3∆.

Corollary 2.12. For any triangleABC , there are the following inequalities:

2 s²−r²−4Rr

≥ X

cyclic

pG(a², b²) (n)≥s²+r²+ 4Rr, (2.62)

s²−2r²−8Rr≥ X

cyclic

r G

(s−a)²,(s−b)²

(n)≥r(4R+r), (2.63)

s²+r²+ 4Rr2

−8s²Rr

4R² ≥ X

cyclic

q

G h²_a, h²_b

(n)≥ 2s²r

R , (2.64)

(4R+r)²−2s² ≥ X

cyclic

q

G r²_a, r_b²

(n)≥s² (2.65)

8R²+r²−s²

8R² ≥ X

cyclic

s G

sin⁴A

2,sin⁴B 2

(n)≥ s²+r²−8Rr

16R² (2.66) and

4 (R+r)²−s²

4R² ≥ X

cyclic

s G

cos⁴ A

2,cos⁴ B 2

(n)≥ s²+ (4R+r)²

8R² . (2.67) Proof. According to Corollary 2.10, we have the inequality

x²+y²+z² ≥ X

cyclic

pG(x², y²) (n)≥xy+yz+zx.

Using the substitutions

(x, y, z) ∈ {(a, b, c),(s−a, s−b, s−c),(ha, hb, hc),(ra, rb, rc),

sin²A

2,sin²B

2,sin²C 2

,

cos² A

2,cos²B

2,cos²C 2

, we deduce the inequalities required.

Corollary 2.13. In any triangle ABC there are the following inequalities:

1

2 3s²−r²−4Rr

≥ X

cyclic

G(a, b) (n)≥s²+r²+ 4Rr, (2.68) 1

2 s²−r²−4Rr

≥ X

cyclic

G(s−a, s−b) (n)≥r(4R+r), (2.69)

s²+r²+ 4Rr2

8R² ≥ X

cyclic

G(h_a, h_b) (n)≥ 2s²r

R (2.70)

and

1 2

h(4R+r)²−s²i

≥ X

cyclic

G(r_a, r_b) (n)≥s². (2.71) Proof. According to Corollary 2.10, we have the inequality

1

2 x²+y²+z²+xy+yz+zx

≥ X

cyclic

G(x, y) (n)≥xy+yz+zx.

Using the substitutions

(x, y, z)∈ {(a, b, c),(s−a, s−b, s−c),(h_a, h_b, h_c),(r_a, r_b, r_c)}, we deduce the inequalities from the statement.

Corollary 2.14. For any triangle ABC there are the following inequalities:

1

4s s²+r²+ 2Rr

≥ Y

cyclic

pG(a, b) (n)≥4sRr, (2.72)

1

2sRr≥ Y

cyclic

pG((s−a),(s−b)) (n)≥sr², (2.73)

s²r s²+r²+ 4Rr

8R² ≥ Y

cyclic

pG(h_a, h_b) (n)≥ 2s²r²

R , (2.74)

1

2s²R ≥ Y

cyclic

pG(r_a, r_b) (n)≥s²r, (2.75)

(2R−r) s²+r²−8Rr

−2Rr²

256R³ ≥

≥ Y

cyclic

s G

sin² A

2,sin² B 2

(n)≥ r²

16R² (2.76)

and

(4R+r)³+s²(2R+r)

256R³ ≥ Y

cyclic

s G

cos² A

2,cos² B 2

(n)≥ s²

16R². (2.77) Proof. According to Corollary 2.10, we have the inequality

1

8(x+y) (y+z) (z+x)≥s Y

cyclic

G(x, y) (n)≥xyz.

Using the substitutions

(x, y, z) ∈ {(a, b, c),(s−a, s−b, s−c),(ha, h_b, hc),(ra, r_b, rc),

sin²A

2,sin²B

2,sin²C 2

,

cos² A

2,cos²B

2,cos²C 2

, we deduce the inequalities required.

Remark 6. From Corollary 2.13, we obtain the inequality 1

4s s²+r²+ 2Rr

≥≥ Y

cyclic

pG(a, b) (n)≥4sRr≥

≥8 Y

cyclic

pG((s−a),(s−b)) (n)≥8sr². (2.78)

REFERENCES

[1] Bencze M. and Minculete N., New refinements of some geometrical inequalities, Octogon Mathematical Magazine, Vol. 16, no.2, 2008.

[2] Botema, O., Djordjevi´c R. Z., Jani´c, R. R., Mitrinovi´c, D. S. and Vasi´c, P.

M., Geometric Inequalities, Gr¨oningen,1969.

[3] Mitrinovi´c, D. S.,Analytic Inequalities, Springer Verlag Berlin, Heidelberg, New York, 1970.

[4] Minculete, N. and Bencze, M., A Generalization of Weitzenb¨ock’s Inequality, Octogon Mathematical Magazine Vol. 16, no.2, 2008.

[5] Minculete, N.,Teoreme ¸si probleme specifice de geometrie, Editura Eurocarpatica, Sfˆantu Gheorghe, 2007 (in Romanian).

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