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ronaldover7

Cho a, b>0 thỏa mãn $\frac{1}{a}+\frac{1}{b}=2$
Chứng minh $\frac{\sqrt{a}}{a^2+b+2b\sqrt{a}}+\frac{\sqrt{b}}{b^2+a+2a\sqrt{b}}$ \leq $\frac{1}{2}$
Ta có: $\frac{\sqrt{a}}{a^2+b+2b\sqrt{a}}+\frac{\sqrt{b}}{b^2+a+2a\sqrt{b}}$ \leq $\frac{\sqrt{a}}{2a\sqrt{b}+2b\sqrt{a}}+ \frac{\sqrt{b}}{2b\sqrt{a}+2a\sqrt{b}}$
=$\frac{\sqrt{a}+\sqrt{b}}{2a\sqrt{b}+2b\sqrt{a}}$=$\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{b}\sqrt{a}(\sqrt{a}+\sqrt{b})}$=$\frac{1}{2\sqrt{b}\sqrt{a}}$ = $\frac{1}{4}$$\frac{2}{\sqrt{b}\sqrt{a}}$ \leq $\frac{1}{4}$ ($\frac{1}{a}+\frac{1}{b}$)=$\frac{2}{4}$ =$\frac{1}{2}$
 
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