# How does the number $3$ help in this (probably Cauchy-based) inequality?

Given $a,b,c>0$ and $a+b+c=3$. Prove that $$\dfrac{a^2+bc}{b+ca}+\dfrac{b^2+ca}{c+ab}+\dfrac{c^2+ab}{a+bc}\ge3$$

Attempt:

Using Cauchy inequality $(a+b\ge2\sqrt{ab})$:

$\dfrac{a^2}{b+ca}+b+ca\ge 2\sqrt{a^2}=2a$

$\dfrac{b^2}{c+ab}+c+ab\ge 2\sqrt{b^2}=2b$

$\dfrac{c^2}{a+bc}+a+bc\ge 2\sqrt{c^2}=2c$

Adding all the inequalities, we have:

$\dfrac{a^2}{b+ca}+\dfrac{b^2}{c+ab}+\dfrac{c^2}{a+bc}+ab+bc+ca+a+b+c\ge2(a+b+c)$

$\Rightarrow\dfrac{a^2}{b+ca}+\dfrac{b^2}{c+ab}+\dfrac{c^2}{a+bc}+ab+bc+ca\ge 3$, but I don't see how that's relevant and I'm stuck here.

By C-S $$\sum_{cyc}\frac{a^2+bc}{b+ca}=\sum_{cyc}\frac{(a^2+bc)^2}{(a^2+bc)(b+ca)}\geq\frac{\left(\sum\limits_{cyc}(a^2+bc)\right)^2}{\sum\limits_{cyc}(a^2+bc)(b+ca)}.$$ Thus, it's enough to prove that $$\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a^2+bc)(b+ca)$$ or $$\sum_{cyc}(a^4+2a^2b^2+a^2b^2+2a^2bc+2a^3b+2a^3c+2a^2bc)\geq$$ $$\geq3\sum_{cyc}(a^2b+a^3c+a^2b+a^2bc)$$ or $$\sum_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2+4a^2bc-2a^3b-2a^2b^2-2a^2bc-3a^3c-3a^2bc)\geq0$$ or $$\sum_{cyc}(a^4-a^3c+a^2b^2-a^2bc)\geq0,$$ which is true by Rearrangement and AM-GM.