[Toán 10] Chứng minh

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vipboycodon

Ta có: $\dfrac{\sqrt{a}}{\sqrt{a+3b}} \le \dfrac{1}{2}(\dfrac{a}{a+b}+\dfrac{a+b}{a+3b})$

$\dfrac{\sqrt{b}}{\sqrt{a+3b}} \le \dfrac{1}{2}(\dfrac{1}{2}+\dfrac{2b}{a+3b})$

$\dfrac{\sqrt{a}}{\sqrt{b+3a}} \le \dfrac{1}{2}(\dfrac{1}{2}+\dfrac{2a}{b+3a})$

$\dfrac{\sqrt{b}}{\sqrt{b+3a}} \le \dfrac{1}{2}(\dfrac{b}{a+b}+\dfrac{a+b}{b+3a})$

Cộng vế với vế ta được: $(\sqrt{a}+\sqrt{b})(\dfrac{1}{\sqrt{a+3b}}+\dfrac{1}{\sqrt{b+3a}}) \le 2$

Dấu "=" xảy ra khi $a = b$
 
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