Question

[dbrby]

cho x,y,z > 0. Cmr: \(\frac{2x}{x^6+y^4}+\frac{2y}{y^6+z^4}+\frac{2z}{z^6+x^4}\le\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)

Answer 1

[The Neil]

\(\frac{1}{x^4}+\frac{1}{y^4}=\frac{x^2}{x^6}+\frac{1}{y^4}\ge\frac{\left(x+1\right)^2}{x^6+y^4}\ge\frac{4x}{x^6+y^4}\)

tương tự

\(\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{4y}{y^6+z^4}\);

\(\frac{1}{z^4}+\frac{1}{x^4}\ge\frac{4z}{z^6+x^4}\);

cộng vế với vế => đpcm

Dấu "=" xảy ra <=> x=y=z=1

Answer 2

[Nguyễn Trần Hoàng]

Cô si