Ta có:
\(x,y>0\)
\(\Rightarrow\left(x+y\right)\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(x^2-y^2\right)\left(x-y\right)\ge0\)
\(\Leftrightarrow x^3-x^2y-xy^2+y^3\ge0\)
\(\Leftrightarrow x^3+y^3\ge x^2y+xy^2=xy\left(x+y\right)\)
Áp dụng:
\(A=\sum\frac{1}{x^3+y^3+1}=\sum\frac{xyz}{x^3+y^3+xyz}\le\sum\frac{xyz}{xy\left(x+y\right)+xyz}=\sum\frac{xyz}{xy\left(x+y+z\right)}=\sum\frac{z}{x+y+z}=1\)
Dấu "=" khi x = y = z = 1
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