[Toán 10] Chứng minh

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vipboycodon

Ta có: aa+3b12(aa+b+a+ba+3b)\dfrac{\sqrt{a}}{\sqrt{a+3b}} \le \dfrac{1}{2}(\dfrac{a}{a+b}+\dfrac{a+b}{a+3b})

ba+3b12(12+2ba+3b)\dfrac{\sqrt{b}}{\sqrt{a+3b}} \le \dfrac{1}{2}(\dfrac{1}{2}+\dfrac{2b}{a+3b})

ab+3a12(12+2ab+3a)\dfrac{\sqrt{a}}{\sqrt{b+3a}} \le \dfrac{1}{2}(\dfrac{1}{2}+\dfrac{2a}{b+3a})

bb+3a12(ba+b+a+bb+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: (a+b)(1a+3b+1b+3a)2(\sqrt{a}+\sqrt{b})(\dfrac{1}{\sqrt{a+3b}}+\dfrac{1}{\sqrt{b+3a}}) \le 2

Dấu "=" xảy ra khi a=ba = b
 
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