Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$ Let $a,b,c>0$ be real numbers such that $a+b+c=3$,how to prove that? : 
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
 A: By AM>GM
$$
\frac{a+b+c}{3}=1 \Rightarrow abc\le 1 \\
\Rightarrow \frac{1}{2ab^2+1} = \frac{1}{2abc\frac{b}{c}+1}\ge\frac{1}{2\frac{b}{c}+1}=\frac{c}{2b+c} \\
S =
\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge\frac{c}{2b+c}+\frac{a}{2c+a}+\frac{b}{2a+b}
$$
$$
\begin{align}
3S-3 & \ge \frac{3c-2b-c}{2b+c}+\frac{3a-2c-a}{2c+a}+\frac{3b-2a-b}{2a+b}\\
& = 2\left(\frac{c-b}{2b+c}+\frac{a-c}{2c+a}+\frac{b-a}{2a+b}\right) \\
& = \frac{2}{D}\left(3ab^2+3bc^2+3ca^2-9abc\right)\\
& = \frac{6abc}{D}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}-3\right) \\
& \ge 0
\end{align}
$$
where for clarity we simply write $D$ for the positive denominator, and the last inequality is again by AM>GM 
$$
1=\left(\frac{b}{c}\cdot\frac{c}{a}\cdot\frac{a}{b}\right)^{1/3}\le\frac{1}{3}\left(\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\right)
$$
Finally $3S-3\ge 0 \Rightarrow S\ge 1$.
A: A solution for straightforward mathematicians. :-)    
It seems the following.
After multiplication of both sides of the inequality by a common denominator and simplification, we reduce the initial inequality to 
$$1+ab^2+bc^2+ca^2\ge 4a^3b^3c^3.$$ 
Since $$ab^2+bc^2+ca^2\ge 3abc$$ it suffices to check that 
$$1+3abc-4(abc)^3\ge 0,$$ 
that is 
$$(1-abc)(2abc+1)^2\ge 0.$$
The last inequality holds because $$abc\le\left(\frac{a+b+c}3\right)^3=1.$$
A: To prove
$$
\frac1{1+2b^2c}+\frac1{1+2c^2a}+\frac1{1+2a^2b}\ge1\tag{1}
$$
subtract $\frac13$ from each term on the left and multiply by $\frac32$:
$$
\frac{1-b^2c}{1+2b^2c}+\frac{1-c^2a}{1+2c^2a}+\frac{1-a^2b}{1+2a^2b}\ge0\tag{2}
$$
Multiplying $(2)$ by $\frac13\left(1+2b^2c\right)\left(1+2c^2a\right)\left(1+2a^2b\right)$ shows that $(1)$ is equivalent to
$$
1+a^2b+b^2c+c^2a-4a^3b^3c^3\ge0\tag{3}
$$
The AM-GM gives
$$
1=\frac{a+b+c}3\ge abc\tag{4}
$$
The AM-GM and $(4)$ yield
$$
\frac{1+a^2b+b^2c+c^2a}4\ge\left(a^3b^3c^3\right)^{1/4}\ge a^3b^3c^3\tag{5}
$$
which is $(3)$.
A: 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
$$\sum_{cyc}\frac{1}{2a^2b+1}=\sum_{cyc}\frac{c^2}{2a^2c^2b+c^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2b^2c+a^2)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2ab+a^2)}=1.$$
