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#1 h Mar 6, 2016, 3:27 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c$ be sides-lengths of a triangle. Prove that $a^2b^2c^2(a+b+c) \geq (a+b-c)(b+c-a)(c+a-b)(a^2b^2+b^2c^2+c^2a^2)$

This post has been edited 1 time. Last edited by Vutienanh00, Mar 6, 2016, 3:38 PM

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Imgoodinmaths_dude

#2 h Mar 6, 2016, 3:34 PM • 2 Y

Y by Adventure10, Mango247

After $a=x+y,b=y+z,c=z+x$ , we have $LHS-RHS=\sum_{sym}{(x^5y^2+2x^5yz+3x^4y^3+9x^4y^2z+14x^3y^3+22x^3y^2z^2-2x^3yz-2x^2y^2z)}\geq 0$

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arqady

#3 h Mar 6, 2016, 3:40 PM • 1 Y

Y by Adventure10

Vutienanh00 wrote:

Let $a,b,c$ be sides-lengths of a triangle. Prove that $a^2b^2c^2(a+b+c) \geq (a+b-c)(b+c-a)(c+a-b)(a^2b^2+b^2c^2+c^2a^2)$

$a^2b^2c^2(a+b+c)-(a+b-c)(b+c-a)(c+a-b)(a^2b^2+b^2c^2+c^2a^2)=$
$=\sum_{cyc}(a^5b^2+a^5c^2-a^4b^3-a^4c^3-a^4b^2c-a^4c^2b+2a^3b^3c)=\sum_{cyc}(a-b)^2a^2b^2(a+b-c)\geq0$ ,
where the last inequality is obviously true even for all non-negatives $a$ , $b$ and $c$ .

This post has been edited 2 times. Last edited by arqady, Mar 6, 2016, 3:41 PM

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#5 h Mar 6, 2016, 4:15 PM • 2 Y

Y by Adventure10, Mango247

Vutienanh00 wrote:

Let $a,b,c$ be sides-lengths of a triangle. Prove that $a^2b^2c^2(a+b+c) \geq (a+b-c)(b+c-a)(c+a-b)(a^2b^2+b^2c^2+c^2a^2)$

Inequality is true for all $a,\,b,\,c$ non-negative numbers.

Indeed, we write inequality as
$(a^2b^2+b^2c^2+c^2a^2)\left[abc-(a+b-c)(b+c-a)(c+a-b)\right] \geqslant abc\left[\sum b^2c^2 -abc(a+b+c)\right],$ equivalent to
$(a^2b^2+b^2c^2+c^2a^2)\left[(a+b-c)(a-b)^2+c(a-c)(b-c)\right] \geqslant abc\left[c^2(a-b)^2+ab(a-c)(b-c)\right],$ or
$\left [ (a^2b^2+b^2c^2+c^2a^2)(a+b-c)-c^3ab\right ](a-b)^2+c^3(a^2+b^2)(a-c)(b-c) \geqslant 0.$ Suppose $a \geqslant b \geqslant c$ then
$\begin{aligned}(a^2b^2+b^2c^2+c^2a^2)(a+b-c)-c^3ab & \geqslant (a^2b^2+b^2c^2+c^2a^2)a-c^3ab \\& = a(a^2b^2+b^2c^2+c^2a^2-bc^3) \geqslant 0.\end{aligned}$ The proof is completed.

This post has been edited 2 times. Last edited by Nguyenhuyen_AG, Mar 6, 2016, 4:17 PM

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suxrob

#6 h Mar 10, 2016, 7:01 PM • 2 Y

Y by Adventure10, Mango247

a, b, c positive numbers, if a + b + c = 3 prove that a / (1 + (b + c)2) + b / (1 + (a + c)2) + c / (1 + (b + a)2) <= 3 x (a2 + b2 +c2) / (a2 + b2 + c2 + 12 x a x b x c)

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suxrob

#7 h Mar 10, 2016, 8:08 PM • 2 Y

Y by Adventure10, Mango247

if a , b, c > 0 and abc = 1 prove that: 1 / sqrt(a3 +2 b3 +6) + 1 / sqrt (b3 + 2 c3 + 6) + 1 / sqrt(c3 + 2 a3 + 6) <= a2 / sqrt(a4 + 4 b + 4 c) + b2 / sqrt(b4 + 4 a + 4 c) + c2 / sqrt(c4 + 4 b + 4 a)

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#8 h Mar 10, 2016, 11:26 PM • 2 Y

Y by Adventure10, Mango247

Vutienanh00 wrote:

Let $a,b,c$ be sides-lengths of a triangle. Prove that $a^2b^2c^2(a+b+c) \geq (a+b-c)(b+c-a)(c+a-b)(a^2b^2+b^2c^2+c^2a^2)$

We have
$LHS-RHS=\sum{(a+b-c)^3c^2(a-b)^2}+(a+b+c)(a-b)^2(b-c)^2(c-a)^2\ge{0}$

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arqady

#9 h Mar 16, 2016, 1:49 AM • 1 Y

Y by Adventure10

Vutienanh00 wrote:

Let $a,b,c$ be sides-lengths of a triangle. Prove that $a^2b^2c^2(a+b+c) \geq (a+b-c)(b+c-a)(c+a-b)(a^2b^2+b^2c^2+c^2a^2)$

The following inequality is also true.
Let $a$ , $b$ and $c$ be non-negative numbers. Prove that:
$2a^2b^2c^2(a+b+c) \geq (a+b-c)(b+c-a)(c+a-b)(a^3b+a^3c+b^3a+b^3c+c^3a+c^3b)$

This post has been edited 2 times. Last edited by arqady, Mar 16, 2016, 2:14 AM

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#10 h Mar 16, 2016, 3:26 AM • 2 Y

Y by Adventure10, Mango247

Vutienanh00 wrote:

Let $a,b,c$ be sides-lengths of a triangle. Prove that $a^2b^2c^2(a+b+c) \geq (a+b-c)(b+c-a)(c+a-b)(a^2b^2+b^2c^2+c^2a^2)$

The following inequality is also true
//cdn.artofproblemsolving.com/images/0/4/c/04ce1a42b8c8c684093f447e3428b5c3f22d9494.jpg

//cdn.artofproblemsolving.com/images/0/4/c/04ce1a42b8c8c684093f447e3428b5c3f22d9494.jpg

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#11 h Mar 16, 2016, 3:56 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

The following inequality is also true.
Let $a$ , $b$ and $c$ be non-negative numbers. Prove that:
$2a^2b^2c^2(a+b+c) \geq (a+b-c)(b+c-a)(c+a-b)(a^3b+a^3c+b^3a+b^3c+c^3a+c^3b)$

We have
$LHS-RHS=\sum{c(a-b)^2(a+b-c)^4}+2(a+b+c)(a-b)^2(a-c)^2(b-c)^2\ge{0}$

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#12 h Mar 16, 2016, 5:12 AM • 2 Y

Y by Adventure10, Mango247

multiply a+b+c on both sides. then apply heron's formula. we get (R(a+b+c))^2>(ab)^2+(bc)^2+(ca)^2. proceeding from here is easy.

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arqady

#15 h Mar 16, 2016, 7:37 AM • 2 Y

Y by Adventure10, Mango247

See also here:
http://www.artofproblemsolving.com/community/c6h559825

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#16 h Mar 16, 2016, 7:44 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

See also here:
http://www.artofproblemsolving.com/community/c6h559825

Thank you very much

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#17 h Mar 16, 2016, 10:23 AM • 2 Y

Y by Adventure10, Mango247

luofangxiang wrote:

The following inequality is also true

http://s5.sinaimg.cn/middle/0018zOAxgy709yRBFSkf4&690

Suppose $a \geqslant b \geqslant c.$ Similar, we write inequality as
$M{{\left( a-b \right)}^{2}}+N\left( a-c \right)\left( b-c \right)\ge 0,$ with
$\begin{aligned} M & = \left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}}+abc \right)\left( a+b-c \right)-abc\left( a+b+c \right) \\ N&=\left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}}+abc \right)c-abc\left( a+b+c \right).\end{aligned}$ We need to prove $M,\,N \geqslant 0.$ Indeed, we have
$\begin{aligned} M & =\left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}}+abc \right)\left( a+b-c \right)-abc\left( a+b+c \right) \\& =\left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}} \right)\left( a+b-c \right)-2abc\cdot c \\& \ge 3abc\cdot a-2abc\cdot c \\& =abc\left( 3a-2c \right)\ge 0.\\ N&=\left( {{a}^{3}}+{{b}^{3}}+{{c}^{3}}+abc \right)c-abc\left( a+b+c \right) \\ & =c\left[ {{a}^{3}}+{{b}^{3}}+{{c}^{3}}+abc-ab\left( a+b+c \right) \right] \\ & =c\left[ {{a}^{3}}+{{b}^{3}}-ab\left( a+b \right) \right] \\ & =c{{\left( a-b \right)}^{2}}\left( a+b \right)\ge 0. \end{aligned}$ The proof is completed.

This post has been edited 1 time. Last edited by Nguyenhuyen_AG, Mar 16, 2016, 10:23 AM

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