Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$

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 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 $$abc\le\left(\frac{a+b+c}3\right)^3=1.$$

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 $(3)$.