Đại 7 nâng cao

E

eye_smile

Ta có:
$\dfrac{x}{2x+y+z}=\dfrac{x}{x+y+x+z}$ \leq $\dfrac{1}{4}(\dfrac{x}{x+y}+\dfrac{x}{x+z})$
TT, ta cũng có: $\dfrac{y}{2y+x+z}$ \leq $\dfrac{1}{4}(\dfrac{y}{x+y}+\dfrac{y}{y+z})$
$\dfrac{z}{2z+x+y}$ \leq $\dfrac{1}{4}(\dfrac{z}{z+x}+\dfrac{z}{z+y})$
Cộng theo vế suy ra đpcm
 
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