<s>[INST] 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? : 
$$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$$
 [/INST] 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 $$abc\le\left(\frac{a+b+c}3\right)^3=1.$$
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