Hpt vip day!!!

T

tuyn

[TEX]\left{\begin{\sqrt{x^2+y^2}+\sqrt{2xy}=8\sqrt{2}(1)}\\{\sqrt{x}+\sqrt{y}=4}[/TEX]
Ta có [TEX]\sqrt{x^2+y^2} \geq \frac{x+y}{\sqrt{2}}[/TEX] \Rightarrow [TEX]VT(1) \geq \frac{x+y}{\sqrt{2}}+\sqrt{2xy}=\frac{x+y+2\sqrt{xy}}{\sqrt{2}}=\frac{(\sqrt{x}+\sqrt{y})^2}{\sqrt{2}}=8\sqrt{2}[/TEX] Dấu ''='' xảy ra khi x=y \Rightarrow x=y=4
 
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