[Toán 9] $a\sqrt{b-1} +b\sqrt{a-1}\le ab$

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thuvan_98

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vansang02121998

Xí luôn bài này

Áp dụng Cauchy

$b-1+1 \ge 2\sqrt{b-1}$

$\Leftrightarrow \dfrac{b}{2} \ge \sqrt{b-1}$

$\Leftrightarrow a\sqrt{b-1} \le \dfrac{ab}{2}$

Tương tự, $b\sqrt{a-1} \le \dfrac{ab}{2}$

$\Rightarrow a\sqrt{b-1}+b\sqrt{a-1} \le ab$
 
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