chứng minh

N

nguyenbahiep1

[laTEX]\frac{\sqrt{a+b}}{\sqrt{2}} \geq \frac{\sqrt{a}+\sqrt{b}}{2} \\ \\ \Leftrightarrow \frac{a+b}{2} \geq \frac{a+b+2\sqrt{ab}}{4} \\ \\ 2(a+b) = a+b + 2\sqrt{ab} \\ \\ a - 2\sqrt{ab}+b \geq 0 \\ \\ (\sqrt{a}-\sqrt{b})^2 \geq 0 \Rightarrow dpcm \\ \\ dau-bang-xay-ra: a = b [/laTEX]
 
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