Ta có:
[tex]\frac{x^4-y^4}{(x^2+y^2)(x+y)} =\frac{(x^2+y^2)(x^2-y^2)(x+y)}{(x^2+y^2)(x+y)}=x-y[/tex]
Tương tự: [tex]\frac{y^4-z^4}{(y^2+z^2)(y+z)}=y-z;\frac{z^4-x^4}{(z^2+x^2)(z+x)}=z-x[/tex]
=> [tex][tex]\frac{x^4-y^4}{(x^2+y^2)(x+y)}+\frac{y^4-z^4}{(y^2+z^2)(y+z)}+\frac{z^4-x^4}{(z^2+x^2)(z+x)}=0.
=> 2P= \frac{x^4+y^4}{(x^2+y^2)(x+y) + \frac{y^4+z^4}{(y^2+z^2)(y+z)} + \frac{z^4+x^4}{(z^2+x^2)(z+x)}.
mà theo bất đẳng thức Cauchy, ta có:
[tex]x^4+y^4\geq \frac{(x^2+y^2)^2}{2} =>\frac{x^4+y^4}{(x^2+y^2)(x+y)}\geq \frac{x^2+y^2}{2(x+y)}\geq \frac{x+y}{4}[/tex]
Tương tự:\frac{y^4+z^4)}{(y^2+z^2)(y+z)\geq \frac{y+z}{4}; \frac{z^4+x^4)}{(z^2+x^2)(z+x)\geq \frac{z+x}{4}
=> 2P\geq \frac{x+y+z}{2} =>P\geq \frac{1}{4}
Dấu bằng xảy ra <=> x=y=z=1/3[/tex][/tex]