[Toán 9] Chứng minh

A

angleofdarkness

$\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}$

$=\dfrac{x}{(x+y)+(z+x)}+\dfrac{y}{(x+y)+(y+z)}$$+\dfrac{z}{(z+x)+(y+z)}$

\leq $\dfrac{x}{4(x+y)}+\dfrac{x}{4(z+x)}+\dfrac{y}{4(x+y)}+\dfrac{y}{4(y+z)}+\dfrac{z}{4(z+x)}+\dfrac{z}{4(y+z)}$

$=\dfrac{x+y}{4(x+y)}+\dfrac{y+z}{4(y+z)}+\dfrac{z+x}{4(z+x)}$

$=\dfrac{3}{4}.$
 
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