- $3(x^{4}+y^{4}+z^{4})\geq (x^{2}+y^{2}+z^{2})^{2}\geq 9\Rightarrow x^{4}+y^{4}+z^{4}\geq 3$
Xét hiệu: $3(x^{3}+y^{3}+z^{3})-(x^{2}+y^{2}+z^{2})(x+y+z)=(x^{3}+y^{3}-xy(x+y))+(y^{3}+z^{3}-yz(y+z))+(z^{3}+x^{3}-zx(z+x))\geq 0\Rightarrow 3(x^{3}+y^{3}+z^{3})\geqslant(x^{2}+y^{2}+z^{2})(x+y+z)\geqslant 3(x+y+z) \Rightarrow x^{3}+y^{3}+z^{3}\geq x+y+z$
Cộng 2 bđt syt ra ĐPCM