TITLE: Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$
QUESTION [8 upvotes]: Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? : 
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$

REPLY [2 votes]: By AM-GM one has $\,abc\leq\left(\frac{a+b+c}{3}\right)^3=1$. And with the
Cauchy–Bunyakovsky–Schwarz inequality we obtain
$$\sum_{cyc}\frac{1}{2a^2b+1}=\sum_{cyc}\frac{c^2}{2a^2c^2b+c^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2b^2c+a^2)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2ab+a^2)}=1.$$