Prove $\sum_{\mathrm{cyc}} (\frac{a^2}{3bc}+\frac{a(b+c)}{b^2+c^2})\ge 4$

Question

Prove $\sum_{\mathrm{cyc}} (\frac{a^2}{3bc}+\frac{a(b+c)}{b^2+c^2} )\ge 4$, preferably with SOS.

My approach: $\sum_{\mathrm{cyc}} \frac{a^2(b^2+c^2)+3abc(b+c)}{3bc(b^2+c^2)} \ge 4$. Let $f(a,b,c)=\frac{a^2(b^2+c^2)+3abc(b+c)}{3bc(b^2+c^2)}$. We see that it is symmetric over $b,c$, but $f(a,c,c)\ne 0$, so help.

Answer 1

It's wrong.

Try $(a,b,c)=(9,10,10).$

In this case $$\frac{a^3+b^3+c^3}{abc}+\frac{a(b+c)}{b^2+c^2}-4=-\frac{61}{900}<0.$$ The following inequality is true already.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^3+b^3+c^3}{abc}+\left(\frac{2bc}{b^2+c^2}\right)^4\geq4.$$ Your new inequality is true by BW.

Indeed, SOS helps here.

We need to prove that $$\frac{a^3+b^3+c^3}{3abc}-1\geq\sum_{cyc}\left(1-\frac{a(b+c)}{b^2+c^2}\right)$$ or $$\frac{(a+b+c)\sum\limits_{cyc}(a-b)^2}{6abc}\geq\sum_{cyc}\frac{c(c-a)-b(a-b)}{b^2+c^2}$$ or $$\frac{(a+b+c)\sum\limits_{cyc}(a-b)^2}{6abc}\geq\sum_{cyc}(a-b)\left(\frac{a}{a^2+c^2}-\frac{b}{b^2+c^2}\right)$$ or $$\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{abc}+\frac{6(ab-c^2)}{(a^2+c^2)(b^2+c^2)}\right)\geq0.$$ Now, let $a\geq b \geq c$.

Thus, $$\frac{a+b+c}{abc}+\frac{6(ab-c^2)}{(a^2+c^2)(b^2+c^2)}\geq0$$ and by AM-GM and C-S $$\frac{a+b+c}{abc}+\frac{6(ac-b^2)}{(a^2+b^2)(b^2+c^2)}\geq\frac{a+b+c}{abc}-\frac{6b^2}{(b^2+a^2)(b^2+c^2)}\geq$$ $$\geq\frac{b+2\sqrt{ac}}{abc}-\frac{6b^2}{(b^2+ac)^2}\geq0$$ because after substitution $b=x\sqrt{ac}$ the last inequality it's $$\frac{x+2}{x}-\frac{6x^2}{(x^2+1)^2}\geq0$$ or $$x^5+2x^4-4x^3+4x^2+x+2\geq0,$$ which is obvious.

Id est, by AM-GM we obtain: $$\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{abc}+\frac{6(ab-c^2)}{(a^2+c^2)(b^2+c^2)}\right)\geq$$ $$\geq(a-c)^2\left(\tfrac{a+b+c}{abc}+\tfrac{6(ac-b^2)}{(a^2+b^2)(b^2+c^2)}\right)+(b-c)^2\left(\tfrac{a+b+c}{abc}+\tfrac{6(bc-a^2)}{(a^2+b^2)(a^2+c^2)}\right)\geq$$ $$\geq(b-c)^2\left(\tfrac{a+b+c}{abc}-\tfrac{6b^2}{(a^2+b^2)(b^2+c^2)}\right)+(b-c)^2\left(\tfrac{a+b+c}{abc}-\tfrac{6a^2}{(a^2+b^2)(a^2+c^2)}\right)=$$ $$=2(b-c)^2\left(\frac{a+b+c}{abc}-\frac{3(2a^2b^2+a^2c^2+b^2c^2)}{\prod\limits_{cyc}(a^2+b^2)}\right)\geq$$ $$\geq2(b-c)^2\left(\frac{a+b+c}{abc}-\frac{8(a^2b^2+a^2c^2+b^2c^2)}{\prod\limits_{cyc}(a^2+b^2)}\right)\geq$$ $$\geq2(b-c)^2\left(\frac{a+b+c}{abc}-\frac{8(a^2b^2+a^2c^2+b^2c^2)}{\frac{8}{9}(a^2+b^2+c^2)(a^2b^2+a^2c^2+b^2c^2)}\right)=$$ $$=\frac{(b-c)^2((a+b+c)(a^2+b^2+c^2)-9abc)}{abc(a^2+b^2+c^2)}\geq0.$$