truonghoangvu11

$\frac{2x^{2}}{1+x^{2}}=2-\frac{2}{1+x^{2}} ; \frac{2y^{2}}{1+y^{2}}=2-\frac{2}{1+y^{2}} ; \frac{2z^{2}}{1+z^{2}}=2-\frac{2}{1+z^{2}}$

Cần c/m: $\frac{2}{1+x^{2}}+\frac{2}{1+y^{2}}+\frac{2}{1+z^{2}}\geq 3$

Theo BCS ta có: $\frac{2}{1+x^{2}}+\frac{2}{1+y^{2}}+\frac{2}{1+z^{2}}\geq \frac{2.9}{3+x^{2}+y^{2}+z^{2}}\geq \frac{18}{3+\sqrt{3.(x^{4}+y^{4}+z^{4})}}=\frac{18}{3+\sqrt{3.3}}=3$ (đpcm)

Dấu = xảy ra khi x=y=z=1