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[adu]

Cho 3 số dương a,b,c thỏa mãn a+b+c=3. cmr : $\sum \frac{a}{a^{2}+9}\leq \frac{3}{10}$

[dinhvu]

$a^2+1\geq 2a\Rightarrow \sum \frac{a}{a^2+9}\leq \sum \frac{a}{2a+8}$
Ta sẽ CM $\sum \frac{a}{2a+8}\leq \frac{3}{10}\Leftrightarrow \sum (\frac{1}{2}-\frac{a}{2a+8})\geq \frac{6}{5}\\ \Leftrightarrow \sum \frac{8}{2(2a+8)}\geq \frac{6}{5}\\ \sum \frac{8}{2(2a+8)}+\sum \frac{2a+8}{25}\geq \frac{12}{5}\\$

Mà $a+b+c=3$ nên $\sum \frac{2a+8}{25}=\frac{6}{5}$$\Rightarrow \sum \frac{a}{a^2+9}\leq \frac{3}{10}$

Edited by dinhvu, 01-10-2023 - 15:47.