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kagali · 1392/5/15

بافرضx,y>0وx+y=1ثابت کنید:
[FONT=&quot]x/√(1-x)+y/√(1-y)≥√2[/FONT]

mrbayat · 1392/5/17

پاسخ : جبر

$\begin{Bmatrix} (\frac{x}{\sqrt{y}} + \frac{y}{\sqrt{x}})(x\sqrt{y}+y\sqrt{x})\geqslant (x+y)^2 = 1 \\ \begin{Bmatrix} A := x\sqrt{y}+y\sqrt{x} = \sqrt{xy}(\sqrt{x}+\sqrt{y})\\ \sqrt{xy}\leq \frac{x+y}{2}=\frac{1}{2}\\ (\sqrt{x}+\sqrt{y})^2 \leq (1+1)(x+y)=2 \end{Bmatrix} \Rightarrow A\leqslant \frac{1}{\sqrt{2}} \end{Bmatrix}\Rightarrow \frac{x}{\sqrt{y}} + \frac{y}{\sqrt{x}}\geqslant \sqrt{2}$

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kagali

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mrbayat

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