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