3. Cho a,b $>0. CM: (\sqrt{a}+\sqrt{b})(\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3a}}) \leq 2$.
Áp dụng bđt AM-GM ta có:
$\sqrt{\frac{a}{a+3b}}=\sqrt{\frac{a(a+b)}{(a+3b)(a+b)}} \leq \frac{1}{2}(\frac{a}{a+b}+\frac{a+b}{a+3b})$
$\sqrt{\frac{b}{a+3b}}\leq \frac{1}{2}(\frac{2b}{a+3b}+\frac{1}{2})$
$\rightarrow \frac{1}{\sqrt{a+3b}}.(\sqrt{a}+\sqrt{b}) \leq \frac{1}{2}.(\frac{a}{a+b}+\frac{a+b+2b}{a+3b}+\frac{1}{2})=\frac{3}{4}+\frac{a}{2(a+b)}$
Tương tự $\frac{1}{\sqrt{b+3a}}.(\sqrt{a}+\sqrt{b}) \leq \frac{3}{4}+\frac{b}{2(a+b)}$
$\rightarrow (\sqrt{a}+\sqrt{b})(\frac{1}{\sqrt{a+3b}}+\frac{1}{\sqrt{b+3a}}) \leq \frac{6}{4}+\frac{1}{2}(\frac{a}{a+b}+\frac{b}{a+b}) = \frac{6}{4}+\frac{1}{2}=2$
Chứng minh hoàn tất.Đẳng thức xảy ra khi $a=b$