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hacbachvothuong

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hacbachvothuong

#1 h Mar 29, 2025, 10:10 AM

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Let $a, b, c$ be positive real numbers such that $ab+bc+ca=3$ . Prove that:
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\ge1$

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sqing

#2 h Mar 29, 2025, 12:08 PM • 1 Y

Y by hacbachvothuong

Let $a,b,c>0$ and $abc=1$ . Prove that
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\geq 1$ Let $a,b,c>0$ and $ab+bc+ca+abc=4$ . Prove that
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\geq 1$

This post has been edited 2 times. Last edited by sqing, Mar 29, 2025, 12:16 PM

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arqady

#3 h Mar 29, 2025, 12:09 PM • 1 Y

Y by hacbachvothuong

hacbachvothuong wrote:

Let $a, b, c$ be positive real numbers such that $ab+bc+ca=3$ . Prove that:
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\ge1$

it's equivalent to $f(w^3)\geq0,$ where $f$ decreases.
Thus, by $uvw$ it's enough to check the case $b=a$ and $c=\frac{3-a^2}{2a},$ which gives $(a-1)^2a^3(a^2+6a+3)\geq0.$

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ektorasmiliotis

107 posts

ektorasmiliotis

#4 h Mar 30, 2025, 10:35 PM

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sqing wrote:

Let $a,b,c>0$ and $abc=1$ . Prove that
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\geq 1$ Let $a,b,c>0$ and $ab+bc+ca+abc=4$ . Prove that
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\geq 1$

i) andreescu and (x+y+z)^2>=3(xy+yz+zx)

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arqady

#5 h Mar 31, 2025, 7:33 PM

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ektorasmiliotis wrote:

sqing wrote:

Let $a,b,c>0$ and $abc=1$ . Prove that
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\geq 1$ Let $a,b,c>0$ and $ab+bc+ca+abc=4$ . Prove that
$\frac{a^2}{a^2+b+c}+\frac{b^2}{b^2+c+a}+\frac{c^2}{c^2+a+b}\geq 1$

i) andreescu and (x+y+z)^2>=3(xy+yz+zx)

Are you sure?

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ektorasmiliotis

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ektorasmiliotis

#7 h Mar 31, 2025, 10:25 PM

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I accidentally used the condition from the first post at some point...thanks for pointing it out. I'll look at it again later

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