Interesting inequality

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sqing

#1 h Mar 23, 2025, 2:12 PM

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Let $a,b,c\geq 0,(ab+c^2)(ac+b^2)\neq 0$ and $a+b+c=3 .$ Prove that
$\frac{1}{ab+c^2}+\frac{1}{ac+b^2} \geq\frac{3}{4}$ $\frac{1}{ab+2c^2}+\frac{1}{ac+2b^2} \geq\frac{4}{9}$

This post has been edited 1 time. Last edited by sqing, Mar 23, 2025, 2:14 PM

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sqing

#2 h Mar 23, 2025, 2:15 PM

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Let $a,b,c\geq 0$ and $a+b+c=3 .$ Prove that
$\frac{1}{a+ b+ c^2}+\frac{1}{a +c+ b^2} \geq\frac{4}{9}$ $\frac{1}{a+ b+ 2c^2}+\frac{1}{a +c+ 2 b^2} \geq\frac{1}{3}$ $\frac{1}{ab+2c^2+3}+\frac{1}{ac+2b^2+3} \geq\frac{4}{15}$

This post has been edited 1 time. Last edited by sqing, Mar 23, 2025, 2:25 PM

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sqing

#4 h Mar 23, 2025, 2:35 PM

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Let $a,b,c\geq 0$ and $a^2+b+c=1 .$ Prove that
$\frac{1}{2a+ b+ c^2}+\frac{1}{2a +c+ b^2} \geq1$

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sqing

#5 h Mar 24, 2025, 6:34 AM

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sqing wrote:

Let $a,b,c\geq 0$ and $a^2+b+c= \frac{1}{4} .$ Prove that $\frac{1}{2a+ b+ c^2}+\frac{1}{2a +c+ b^2} \geq2$

Solution of DAVROS:
$\frac{1}{2a+ b+ c^2}+\frac{1}{2a +c+ b^2} \ge \frac4{4a+b+c+b^2+c^2} \ge \frac4{4a+b+c+(b+c)^2} = \frac{64}{16a^4 - 24a^2 + 64 a + 5}$ for $a\le\frac12$

$\text{LHS} \ge \frac{64}{32- (1-2a)(8 a^3 + 4 a^2 - 10 a + 27)} \ge 2$ @ $a=\frac12, b=c=0$

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#6 h Monday at 12:04 PM

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let : $b\geq a\geq c$
$\frac{1}{ab+c^2}+\frac{1}{ac+b^2} \geq \frac{1}{a(b+c)}+\frac{1}{(b+c)^2}$ $\frac{1}{a(b+c)}+\frac{1}{(b+c)^2} = \frac{1}{(3-b-c)(b+c)}+\frac{1}{(b+c)^2} \geq \frac{3}{4}$ equality cases : $(a,b,c)=(1,2,0)=(1,0,2)$

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sqing

#7 h Monday at 12:17 PM

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Nice.Thanks.

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