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Interesting inequality
sqing   5
N Monday at 12:17 PM by sqing
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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
$$ \frac{1}{ab+c^2}+\frac{1}{ac+b^2} \geq\frac{3}{4} $$$$ \frac{1}{ab+2c^2}+\frac{1}{ac+2b^2} \geq\frac{4}{9} $$
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sqing
Mar 23, 2025
sqing
Monday at 12:17 PM
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sqing
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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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#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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#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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#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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SunnyEvan
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SunnyEvan
#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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#7 h  Monday at 12:17 PM
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Nice.Thanks.
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