Chứng minh:
Ta có: $xy +yz +zx \le x^2 +y^2 +z^2$ với mọi $x,y,z$ không âm
\Rightarrow $xy +yz +zx \le 3$
Lại có: $3 =x^2 +y^2 +z^2 \ge \dfrac{1}{3}(a +b +c)^2$
\Leftrightarrow $x +y +z \le 3$
Theo giả thiết, ta có: $(x +y +z)^2 \ge 3(xy +yz +zx)$ với mọi x,y,z không âm
\Rightarrow $x +y +z \ge \sqrt{3(xy +yz +zx)}$
\Rightarrow $\dfrac{1}{x +y +z} \le \dfrac{1}{\sqrt{3(xy +yz +zx)}}$
$\le \dfrac{1}{\sqrt{xy} +\sqrt{yz} +\sqrt{zx}}$
$\le \dfrac{1}{9}(\dfrac{1}{\sqrt{xy}} +\dfrac{1}{\sqrt{yz}} +\dfrac{1}{\sqrt{zx}})$
\Rightarrow $A= xy +yz +zx +\dfrac{1}{x +y +z} \le xy +yz +zx +\dfrac{1}{9}(\dfrac{1}{\sqrt{xy}} +\dfrac{1}{\sqrt{yz}} +\dfrac{1}{\sqrt{zx}}$
\Rightarrow $A \le (\dfrac{xy}{9} +\dfrac{1}{9\sqrt{xy}}) +(\dfrac{yz}{9} +\dfrac{1}{9\sqrt{yz}}) +(\dfrac{zx}{9} +\dfrac{1}{9\sqrt{zx}}) + \dfrac{8(xy +yz +zx)}{9}$
\Rightarrow $A \le \dfrac{2}{9}({4}\sqrt{xy} +{4}\sqrt{yz} +{4}\sqrt{zx}) +\frac{8}{3}$
\Rightarrow $A \le \dfrac{2}{9}(\sqrt{x} +\sqrt{y} +\sqrt{z}) +\dfrac{8}{3}$
\Rightarrow $A \le \dfrac{2}{9}\sqrt{3(x +y +z)} + \dfrac{8}{3} \le \dfrac{2}{3} +\dfrac{8}{3} \le \dfrac{10}{3}$
Dấu ''='' xảy ra \Leftrightarrow $x =y =z =1$
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