Ta có: $(x+y+z+t)^2 \\ = (x+z)^2+(y+t)^2+2(x+z)(y+t) \\ \ge \dfrac12(x+y+z+t)^2+2(xy+yz+zt+tx) $
\Leftrightarrow $\dfrac12(x+y+z+t)^2 \ge 2(xy+yz+zt+tx)$
\Leftrightarrow $xy+yz+zt+tx \le 4$
Tương tự:
$\dfrac12(xy+yz+zt+tx)^2 \ge 2(xy^2z+yz^2t+zt^2x+x^2yt)$
\Rightarrow $8 \ge 2(xy^2z+yz^2t+zt^2x+x^2yt)$
\Leftrightarrow $4 \ge xy^2z+yz^2t+zt^2x+x^2yt$
\Leftrightarrow $(x+xy^2z)+(y+yz^2t)+(z+zt^2x)+(t+x^2yt) \le 8$
\Leftrightarrow $\dfrac{16}{(x+xy^2z)+(y+yz^2t)+(z+zt^2x)+(t+x^2yt)} \ge 2$
$P=\dfrac{x^2}{x+xy^2z}+\dfrac{y^2}{y+yz^2t}+ \dfrac{z^2}{z+zt^2x}+\dfrac{t^2}{t+tx^2y} \ge \dfrac{(x+y+z+t)^2}{(x+xy^2z)+(y+yz^2t)+(z+zt^2x)+(t+x^2yt)} \ge 2$