Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{2x+y+z}=\frac{1}{(x+y)+(x+z)}\leq \frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)
\(\Rightarrow \frac{x}{2x+y+z}\leq \frac{1}{4}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\)
Tương tự:
\(\frac{y}{2y+x+z}\leq \frac{1}{4}\left(\frac{y}{y+z}+\frac{y}{y+x}\right)\)
\(\frac{z}{2z+x+y}\leq \frac{1}{4}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)\)
Cộng theo vế:
\(D\leq \frac{1}{4}\left(\frac{x+y}{x+y}+\frac{y+z}{y+z}+\frac{z+x}{z+x}\right)=\frac{3}{4}\) (dpcm)
Dấu bằng xảy ra khi $x=y=z$