Stuck on this cyclic 3-variables-inequality with constraint

Question

Problem Statement:

For $a,b,c>0$ and $a+b+c=3$, I'd want to prove that $$\frac{1}{a^2+b+c}+\frac{1}{b^2+a+c} +\frac {1}{c^2+a+b} \leq1 .$$

I am a beginner when it comes to inequalities. This problem appeared on my test and I tried everything I knew:

1. AM-GM-HM,
2. Cauchy-Schwarz,
3. homogenization,
4. elimination,
5. properties of quadratic functions,
6. etc ....

but I was not not able to prove it. In many cases I got the upper bound as greater than one, so I knew that my result was very weak compared to the inequality given.

I just need a hint. :-)

Answer 1

By C-S $$\sum_{cyc}\frac{1}{a^2+b+c}=\sum_{cyc}\frac{1+b+c}{(1+b+c)(a^2+b+c)}\leq\sum_{cyc}\frac{1+b+c}{(a+b+c)^2}=1.$$

Answer 2

Also, the Tangent Line method helps.

Indeed, $$1-\sum_{cyc}\frac{1}{a^2+b+c}=\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^2-a+3}\right)=\frac{1}{3}\sum_{cyc}\frac{a^2-a}{a^2-a+3}=$$ $$=\frac{1}{3}\sum_{cyc}\left(\frac{a^2-a}{a^2-a+3}-\frac{a-1}{3}\right)=\frac{1}{9}\sum_{cyc}\frac{(a-1)^2(3-a)}{a^2-a+3}\geq0.$$