B1:ta có$\color{grey}{:\frac{x^2}{x+2y^2}=\frac{x(x+2y^2)-2xy^2}{x+2y^2}=x-\frac{2xy^2}{x+y^2+y^2}\geq x-\frac{2}{3}\sqrt[3]{x^2y^2}}$
$TT...........$
$\color{purple}{P\geq x+y+z-\frac{2}{3}(\sqrt[3]{x^2y^2}+\sqrt[3]{x^2z^2}+\sqrt[3]{y^2z^2})}$
lại có:$\color{red}{\sqrt[3]{x^2y^2}\leq \frac{x^2+y^2+1}{3},TT.......}$
$\color{green}{\Rightarrow P\geq x+y+z-\frac{4}{9}(x^2+y^2+z^2+\frac{3}{2})\geq x+y+z-\frac{4}{9}[\frac{1}{3}(x+y+z)^2+\frac{3}{2}]=1}$