$\sum \frac{1}{\sqrt{3+a}}\leq \sqrt{3\left ( \sum \frac{1}{a+3 } \right )}$
Ta sẽ chứng minh $\sum \frac{1}{3+a}\leq \frac{3}{4}$
Đổi biến $\left ( a,b,c \right )= \left ( \frac{x}{y},\frac{y}{z},\frac{z}{x} \right )$
Đpcm $\Leftrightarrow \sum \frac{y}{x+3y}\leq \frac{3}{4}\Leftrightarrow \sum \frac{3y}{x+3y}\leq \frac{9}{4}\Leftrightarrow \sum \frac{x}{x+3y}\geq \frac{3}{4}$
Điều này luôn đúng do $\sum \frac{x}{x+3y}= \sum \frac{x^{2}}{x^{2}+3xy}\geq \frac{\left ( \sum x \right )^{2}}{\sum x^{2}+3\sum xy}= \frac{\left ( \sum x \right )^{2}}{\left ( \sum x \right )^{2}+\sum xy}\geq \frac{\left ( \sum x \right )^{2}}{\left ( \sum x \right )^{2}+\frac{1}{3}\left ( \sum x \right )^{2}}=\frac{3}{4}$