Cho $a,b,c,d>0$ thỏa mãn: $a+b+c+d=2$. Chứng minh rằng:
$\frac{1}{1+3a^2}+\frac{1}{1+3b^2}+\frac{1}{1+3c^2}+\frac{1}{1+3d^2}\ge \frac{16}{7}$
Ta có : $(\sum \frac{1}{1+3a^{2}})(\sum 1+3a^{2})(1+1+1+1)(1+1+1+1)\geq 256 ( holder)$
Mà : $(\sum \frac{1}{1+3a^{2}})(\sum 1+3a^{2})(1+1+1+1)(1+1+1+1)\geq (\sum \frac{1}{1+3a^{2}})(4+\frac{3(\sum a)^{2}}{4})X16\geq 112(\sum \frac{1}{1+3a^{2}})$
=> $\sum \frac{1}{1+3a^{2}}\geq \frac{16}{7}$( Q.E.D )