Bài Số Tuyển Sinh 10 Khó!

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lmphuc99

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muttay04

$ \frac{{2(a\^2 + 2ab + b\^2)}}{4} + \frac{{a + b}}{4} \ge \frac{{4a\sqrt b + 4b\sqrt a }}{4} \\
\Leftrightarrow 2a\^2 + 4ab + 2b\^2 + a + b \ge 4a\sqrt b + 4b\sqrt a \\
\Leftrightarrow 2a\^2 + 4ab + 4b\^2 + a + a - 4a\sqrt b - 4b\sqrt a \ge 0 \\
\Leftrightarrow (2 + - \sqrt a )\^2 + (a - \sqrt b )\^2 + 4ab - 2a\sqrt b + a + b \\ $
Chứng ming 4ab>$2a\sqrt[2]{b}$
a,b>0 =>đpcm
cái dấu gạch là mũ nha pạn
 
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