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[Matthew James]

Với $a,b,c>0$ thỏa mãn $ab+bc+ca=1$, chứng minh rằng: 

$\frac{a}{1+a^2}+\frac{b}{1+b^2}\leq \frac{1}{\sqrt{1+c^2}}$

Edited by Matthew James, 29-12-2022 - 19:54.

[Sangnguyen3]

$\Leftrightarrow \frac{a+ab^2 + b+ a^2b}{(a+b)^{2}(a+c)(b+c)}\leq \frac{1}{\sqrt{(c+a)(c+b)}}$

$\Leftrightarrow \left (\frac{ab+1}{(a+b)(b+c)(c+a)} \right )^{2}\leq \frac{1}{(c+a)(c+b)}$

$\Leftrightarrow (ab+1)^{2}\leq (a+b)(a+c)(b+a)(b+c)$

$\Leftrightarrow (ab+1)^{2}\leq (a^2+1)(b^2+1)$

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