Cho $x,y,z>0$ thỏa $x^2+y^2+z^2=3$.Chứng minh rằng:
$\frac{1}{{{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}}+\frac{1}{{{\left( {{y}^{2}}+{{z}^{2}} \right)}^{2}}}+\frac{1}{{{\left( {{x}^{2}}+{{z}^{2}} \right)}^{2}}}\ge \frac{3}{32}xyz({{x}^{2}}+{{y}^{2}})({{y}^{2}}+{{z}^{2}})({{x}^{2}}+{{z}^{2}})$
Dễ có $xyz\leqslant 1\Rightarrow$ ta sẽ chứng minh $\sum \frac{1}{(x^2+y^2)^2}\geqslant \frac{3}{32}(x^2+y^2)(y^2+z^2)(z^2+x^2)$
Đặt $(a,b,c)=(x^2+y^2,...)\Rightarrow a+b+c=6$ suy ra $abc\leqslant 8$
BĐT đưa về cm $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geqslant \frac{3.abc}{32}$
Áp dụng BĐT Cô si
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geqslant 3\sqrt[3]{\frac{1}{a^2b^2c^2}}\geqslant 3\sqrt[3]{\frac{1}{64}}=\frac{3}{4}(1)$
$\frac{3.abc}{32}\leqslant \frac{3.8}{32}=\frac{3}{4}(2)$
Từ $(1);(2)$ ta có đpcm