Lời giải:
Áp dụng BĐT Bunhiacopxky + AM- GM:
\((x\sqrt{1-y^2}+y\sqrt{1-z^2}+z\sqrt{1-x^2})^2\leq (x^2+y^2+z^2)(1-y^2+1-z^2+1-x^2)\)
\(=(x^2+y^2+z^2)(3-x^2-y^2-z^2)\leq \left(\frac{x^2+y^2+z^2+3-x^2-y^2-z^2}{2}\right)^2=(\frac{3}{2})^2\)
\(\Rightarrow x\sqrt{1-y^2}+y\sqrt{1-z^2}+z\sqrt{1-x^2}\leq \frac{3}{2}\)
Dấu "=" xảy ra khi \(\frac{x}{\sqrt{1-y^2}}=\frac{y}{\sqrt{1-z^2}}=\frac{z}{\sqrt{1-x^2}}\) và $x^2+y^2+z^2=\frac{3}{2}$
$\Leftrightarrow x=y=z=\frac{1}{\sqrt{2}}$ (do $x,y,z$ dương)
$\Rightarrow P=\frac{3}{4}$