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[yeudiendanlamlam]

Cho $x,y,z>0$. Tìm max của $\frac{2xy+4yz+3zx}{(x+y+z)^2}$

Edited by yeudiendanlamlam, 06-07-2016 - 15:38.

[Baoriven]

Ta có: $P=2ab+4bc+3ca$ với: $a=\frac{x}{x+y+z};b=\frac{y}{x+y+z};c=\frac{z}{x+y+z}; a+b+c=1$

Khi đó: $P=\frac{1}{2}a(b+c)+\frac{3}{2}b(c+a)+\frac{5}{2}c(a+b)$

               $=\frac{9}{8}-\frac{1}{2}[(a-\frac{1}{2})^2+3(b-\frac{1}{2})^2+5(c-\frac{1}{2})^2]$

               $\leq \frac{9}{8}-\frac{(a-\frac{1}{2}+b-\frac{1}{2}+c-\frac{1}{2})^2}{2(1+\frac{1}{3}+\frac{1}{5})}=\frac{24}{23}$

Đẳng thức xảy ra khi: $x=3y-1=5z-2$.