Toán 12 Tìm giá trị lớn nhất

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vansang02121998

$\dfrac{3}{50}(12x+1)-\dfrac{x}{x^2+1}=\dfrac{(3x-1)^2(4x+3)}{50(x^2+1)} \ge 0$

$\Rightarrow \dfrac{x}{x^2+1}+\dfrac{y}{y^2+1}+\dfrac{z}{z^2+1} \le \dfrac{3}{50}(12x+12y+12z+3)=\dfrac{9}{10}$
 
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