Find minimum $\sum\frac{a^2(b+c)}{b^2+bc+c^2}$ when $a^2+b^2+c^2=a+b+c$

Question

Let $a,b,c\ge 0: ab+bc+ca>0$ such that $a^2+b^2+c^2=a+b+c.$ Find minimum $$M=\frac{a^2(b+c)}{b^2+bc+c^2}+\frac{b^2(c+a)}{c^2+ca+a^2}+\frac{c^2(a+b)}{a^2+ab+b^2}$$ When $a=b=c=1$ we get minimum is equal to $2$ By Cauchy-Schwarz $$M\ge \frac{4(ab+bc+ca)^2}{\sum\limits_{cyc}(b^2+bc+c^2)(b+c)}\ge 2$$ $$\iff 2(ab+bc+ca)^2\ge \sum\limits_{cyc}(b^2+bc+c^2)(b+c)$$

What is the rest?

Answer 1

The rest is nothing because you got a wrong inequality. Try $b=c=0$ and $a=1$.

My solution.

We need to prove that $$\sum_{cyc}\frac{a^2(b+c)}{b^2+bc+c^2}\geq\frac{2(a^2+b^2+c^2)}{a+b+c}$$ or $$\sum_{cyc}\left(\frac{a^2(b+c)}{b^2+bc+c^2}-\frac{2a^2}{a+b+c}\right)\geq0$$ or $$\sum_{cyc}\frac{a^2(b(a-b)-c(c-a))}{b^2+bc+c^2}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{a^2b}{b^2+bc+c^2}-\frac{b^2a}{a^2+ac+c^2}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2ab\sum\limits_{cyc}(a^2+ab)}{(a^2+ac+c^2)(b^2+bc+c^2)}\geq0$$ and we are done.

Answer 2

Proof.

Let $a=b=c=1,$ or $a=0$ we obtain $M\ge 2.$

We will prove $$\sum_{cyc}\frac{a^2(b+c)}{b^2+bc+c^2}\ge \frac{2(a^2+b^2+c^2)}{a+b+c}.$$ By using Cauchy-Schwarz$$\sum_{cyc}\left[\frac{a^2(b+c)}{b^2+bc+c^2}+\frac{a^2(b+c)}{ab+bc+ca}\right]\ge 4\sum_{cyc}\frac{a^2(b+c)}{(b+c)(a+b+c)}= 4\sum_{cyc}\frac{a^2}{a+b+c},$$which leads to$$\sum_{cyc}\frac{a^2(b+c)}{b^2+bc+c^2}\ge \frac{2(a^2+b^2+c^2)}{a+b+c}+\frac{(a+b+c)^2-4(ab+bc+ca)}{a+b+c}+\frac{3abc}{ab+bc+ca}.$$ It suffices to prove $$\frac{(a+b+c)^2-4(ab+bc+ca)}{a+b+c}+\frac{3abc}{ab+bc+ca}\ge 0.\tag{1}$$ Now, we can split $(1)$ into two cases

• $(a+b+c)^2-4(ab+bc+ca)>0:$ $(1)$ is obviously true.
• $(a+b+c)^2-4(ab+bc+ca)\le 0:$ By using Schur of third degree \begin{align*} &\frac{(a+b+c)^2-4(ab+bc+ca)}{a+b+c}+\frac{3abc}{ab+bc+ca}\\&\ge [-(a+b+c)^2+4(ab+bc+ca)]\left(\frac{a+b+c}{3(ab+bc+ca)}-\frac{1}{a+b+c}\right)\ge 0. \end{align*}

The desired result follows. Equality holds at $a=b=c=1$ or $abc=0.$