Áp dụng BĐT Cauchy-Schwarz ta có:
\(\dfrac{x}{2x+y+z}=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{y}{2y+x+z}\le\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right);\dfrac{z}{2z+y+x}\le\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{x+y}+\dfrac{y}{y+z}\right)+\dfrac{1}{4}\left(\dfrac{z}{y+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)\)
\(=\dfrac{1}{4}\left(\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}\right)=\dfrac{1}{4}\left(1+1+1\right)=\dfrac{3}{4}\)
