Lời giải:
Theo hệ quả của BĐT AM-GM:
\(x^2+y^2+z^2\geq xy+yz+xz\)
\(\Leftrightarrow (x+y+z)^2\geq 3(xy+yz+xz)\Leftrightarrow xy+yz+xz\leq 3\)
Do đó:
\(P=\sum \frac{xy}{\sqrt{z^2+3}}\leq \sum \frac{xy}{\sqrt{z^2+xy+yz+xz}}\)
\(\Leftrightarrow P\leq \sum \frac{xy}{\sqrt{(z+x)(z+y)}}\) (1)
Áp dụng BĐT AM-GM:
\(\frac{2xy}{\sqrt{(z+x)(z+y)}}\leq \frac{xy}{z+x}+\frac{xy}{z+y}\)
\(\frac{2yz}{\sqrt{(y+x)(x+z)}}\leq \frac{yz}{y+x}+\frac{yz}{x+z}\)
\(\frac{2xz}{\sqrt{(x+y)(y+z)}}\leq \frac{xz}{x+y}+\frac{xz}{z+y}\)
Cộng theo vế:
\(2\sum \frac{xy}{\sqrt{(z+x)(z+y)}}\leq \frac{y(x+z)}{x+z}+\frac{x(y+z)}{y+z}+\frac{z(x+y)}{x+y}\)
\(\Leftrightarrow 2\sum \frac{xy}{\sqrt{(z+y)(z+x)}}\leq x+y+z=3\)
\(\Leftrightarrow \sum \frac{xy}{\sqrt{(z+y)(z+x)}}\leq \frac{3}{2}(2)\)
Từ \((1);(2)\Rightarrow P\leq \frac{3}{2}\Leftrightarrow P_{\max}=\frac{3}{2}\)
Dấu bằng xảy ra khi \(x=y=z=1\)
