若正数$x,y,z$满足$x+y+z=1$,求证:\[\frac{x^4}{z(1-z^2)}+\frac{y^4}{x(1-x^2)}+\frac{z^4}{y(1-y^2)}\geqslant\frac{1}{8}\]
证明：由柯西不等式有\[\frac{x^4}{z(1-z^2)}+\frac{y^4}{x(1-x^2)}+\frac{z^4}{y(1-y^2)}\geqslant\frac{(x^2+y^2+z^2)^2}{x+y+z-x^3-y^3-z^3}=\]
\[\frac{(x^2+y^2+z^2)^2}{1-(x+y+z)(x^3+y^3+z^3)}\geqslant\frac{1/9}{1-1/9}=\frac{1}{8}\]
证明：由柯西不等式有\[\frac{x^4}{z(1-z^2)}+\frac{y^4}{x(1-x^2)}+\frac{z^4}{y(1-y^2)}\geqslant\frac{(x^2+y^2+z^2)^2}{x+y+z-x^3-y^3-z^3}=\]
\[\frac{(x^2+y^2+z^2)^2}{1-(x+y+z)(x^3+y^3+z^3)}\geqslant\frac{1/9}{1-1/9}=\frac{1}{8}\]
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