2 var inequalities

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sqing

#1 h Apr 23, 2025, 1:13 PM

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Let $a,b> 0$ and $a+b\leq 2ab .$ Prove that
$\frac{ a + b }{ a^2(1+ b^2)} \leq \sqrt 5-1$ $\frac{ a +ab+ b }{ a^2(1+ b^2)} \leq \frac{3(\sqrt5-1)}{2}$ $\frac{ a +a^2b^2+ b }{ a^2(1+ b^2)} \leq2$ Solution:
$a\ge\frac{b}{2b-1}, b>\frac12$ and $\frac{ a +a^2b^2+ b }{ a^2(1+ b^2)} \le\frac{2ab+a^2b^2}{a^2(1+b^2)}=1+\frac{2b-a}{a(1+b^2)} \le 1+\frac{4b-3}{b^2+1}$

Assume $u=4b-3>0$ then $\frac{ a +a^2b^2+ b }{ a^2(1+ b^2)} \le 1+\frac{16u}{u^2+6u+25} =2+ \frac{16}{6+u+\frac{25}u} \le 3$
Equalityholds when $a=\frac{2}{3},b=2.$

This post has been edited 2 times. Last edited by sqing, Apr 24, 2025, 2:54 PM

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sqing

#2 h Apr 23, 2025, 1:19 PM

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Let $a,b> 0$ and $a+b\leq 2ab .$ Prove that
$\frac{ a + b }{ a^2(1+ 2b^2)} \leq \sqrt{3}-1$ $\frac{ a +ab+ b }{ a^2(1+ 2b^2)} \leq \frac{3(\sqrt{3}-1)}{2}$ $\frac{ a +a^2b^2+ b }{ a^2(1+2 b^2)} \leq\frac{\sqrt{57}-3 }{4}$ $\frac{ a + b }{ a^2(2+ b^2)} \leq 1$ $\frac{ a +ab+ b }{ a^2(2+ b^2)} \leq \frac{3 }{2}$ $\frac{ a +a^2b^2+ b }{ a^2(2+b^2)} \leq \sqrt{3}$

This post has been edited 2 times. Last edited by sqing, Apr 24, 2025, 11:08 AM

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sqing

#3 h Apr 24, 2025, 8:57 AM

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sqing wrote:

Let $a,b> 0$ and $a+b\leq 2ab .$ Prove that $\frac{ a +2a^2b^2+ b }{ a^2(1+ b^2)}\leq 2\sqrt2$

Solution of DAVROS:
$a\ge\frac{b}{2b-1}, b>\frac12$ and $\textbf{LHS} \le\frac{2ab+2a^2b^2}{a^2(1+b^2)}=2+\frac{2(b-a)}{a(1+b^2)} \le 2+\frac{4(b-1)}{b^2+1}$

For max can assume $u=b-1>0$ then $\text{LHS} \le 2+\frac{4u}{u^2+2u+2} =2+ \frac4{2+u+\frac2u} \le 2\sqrt2$

Equality is possible at $b-1=\sqrt2, a=\frac{b}{2b-1}\implies b=\sqrt2+1, a= \frac{3+\sqrt2}7$

This post has been edited 1 time. Last edited by sqing, Apr 24, 2025, 11:10 AM

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sqing

#4 h Apr 24, 2025, 3:03 PM

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Let $a,b> 0$ and $a+b\leq 2ab .$ Prove that
$\frac{ a - ab+ b }{ a^2(2+ b^2)} \leq \frac{1}{2}$ $\frac{ a - ab+ b }{ a^2(1+ b^2)}\leq \frac{\sqrt 5-1}{2}$ $\frac{ a - a^2b^2+ b }{ a^2(2+ b^2)}\leq \sqrt {2}-1$ $\frac{ a - a^2b^2+ b }{ a^2(2+3b^2)}\leq \frac{\sqrt{7}-2}{3}$

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