prove that $\sum_{cyc}\frac{a(b^2+c^2)}{a^2+bc}\ge 3$

Question

if $a,b,c$ are positive prove $$\sum_{cyc}\frac{a(b^2+c^2)}{a^2+bc}\ge 3$$ given $ab+bc+ca=3$

My try:using AM-GM and Titu's lemma :

$$\sum_{cyc}\frac{a(b^2+c^2)}{a^2+bc}\ge 2\sum_{cyc}\frac{abc}{a^2+bc}\ge \frac{18abc}{a^2+b^2+c^2+ab+bc+ca}$$

But $a^2+b^2+c^2+ab+bc+ca\le 6abc$ need not be true!

I am stuck

Answer 1

Using known inequality $(x+y+z)^2 \geqslant 3(xy+yz+zx),$ we have $$\left[\sum \frac{a(b^2+c^2)}{a^2+bc}\right] ^2 \geqslant 3 \sum \frac{ab(a^2+c^2)(b^2+c^2)}{(a^2+bc)(b^2+ca)}.$$ Therefore, we need to prove $$\sum \frac{ab(a^2+c^2)(b^2+c^2)}{(a^2+bc)(b^2+ca)} \geqslant ab+bc+ca,$$ equivalent to $$abc\left[\frac{a(a-b)(a-c)}{(b^2+ca)(c^2+ab)}+\frac{b(b-c)(b-a)}{(c^2+ab)(a^2+bc)}+\frac{c(c-a)(c-b)}{(a^2+bc)(b^2+ca)}\right] \geqslant 0.$$ Suppose $a \geqslant b \geqslant c$ then $$\frac{c(c-a)(c-b)}{(a^2+bc)(b^2+ca)} \geqslant 0,$$ and $$\frac{a(a-b)(a-c)}{(b^2+ca)(c^2+ab)}+\frac{b(b-c)(b-a)}{(c^2+ab)(a^2+bc)} = \frac{(a^2+b^2)(a-b)^2(a+b-c)}{(a^2+bc)(b^2+ca)(c^2+ab)} \geqslant 0.$$ The proof is completed.

Answer 2

Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v>0$ and $abc=w^3$.

Thus, our inequality is equivalent to $f(w^3)\leq0$, where $$f(w^3)= (10u+8v)w^6+9(6u^4-7u^2v^2+3u^3v-6uv^3-v^4)w^3-81u^3v^4+81uv^6+27v^7.$$ Since $10u+8v>0$, we see that $f$ is a convex function.

Thus, it's enough to prove our inequality for an extreme value of $w^3$,

which by $uvw$ happens in the following cases:

1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$.

Thus, we need to prove that: $$\frac{3b}{a^2}+\frac{3a}{b^2}\geq3,$$ where $ab=3,$ which is true by C-S: $$\frac{3b}{a^2}+\frac{3a}{b^2}=3\left(\frac{b^2}{a^2b}+\frac{a^2}{b^2a}\right)\geq\frac{3(a+b)^2}{ab(a+b)}=a+b\geq2\sqrt{ab}=2\sqrt3>3.$$ 2. Two variables are equal.

Let $b=a$.

Thus, $$c=\frac{3-a^2}{2a}$$ and we need to prove that: $$(a-1)^2(7a^6+9a^5-9a^4-36a^3-9a^2+27a+27)\geq0,$$ which is true.

Can you end it now?

Answer 3

Another way. $$\sum_{cyc}\frac{a(b^2+c^2)}{a^2+bc}=\sqrt{\sum_{cyc}\frac{a^2(b^2+c^2)^2}{(a^2+bc)^2}+2\sum_{cyc}\frac{ab(a^2+c^2)(b^2+c^2)}{(a^2+bc)(b^2+ac)}}\geq$$ $$\geq\sqrt{3\sum_{cyc}\frac{ab(a^2+c^2)(b^2+c^2)}{(a^2+bc)(b^2+ac)}}.$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{ab(a^2+c^2)(b^2+c^2)}{(a^2+bc)(b^2+ac)}\geq ab+ac+bc$$ or $$abc\sum_{cyc}(a^5-a^4b-a^4c+2a^3bc-a^2b^2c)\geq0,$$ which is true by Schur: $$\sum_{cyc}(a^5-a^4b-a^4c+a^3bc)\geq0$$ and Muirhead: $$\sum_{cyc}(a^3bc-a^2b^2c)\geq0.$$

Answer 4

Another way.

Since for positives $x$, $y$, $z$, $a$, $b$ and $c$ we have $$x(b+c)+y(c+a)+z(a+b)\geq2\sqrt{(xy+xz+yz)(ab+ac+bc)}$$ (the proof see here: Bordered Hessian matrix to find a minimum of the function ),

for $x=\frac{a(b^2+c^2)}{b(a^2+c^2)+c(a^2+b^2)}$, $y=\frac{b(a^2+c^2)}{a(b^2+c^2)+c(a^2+b^2)}$ and $z=\frac{c(a^2+b^2)}{a(b^2+c^2)+b(a^2+c^2)}$ we obtain: $$\sum_{cyc}\frac{a(b^2+c^2)}{a^2+bc}=\sum_{cyc}\frac{a(b^2+c^2)(b+c)}{(a^2+bc)(b+c)}=\sum_{cyc}\frac{a(b^2+c^2)(b+c)}{b(a^2+c^2)+c(a^2+b^2)}=$$ $$=x(b+c)+y(a+c)+z(a+b)\geq\sqrt{4(xy+xz+yz)(ab+ac+bc)}.$$ Id est, it's enough to prove that $$4(xy+xz+yz)\geq3.$$ Now, let $a(b^2+c^2)=p$, $b(a^2+c^2)=q$ and $c(a^2+b^2)=r$.

Thus, we need to prove that: $$4\sum_{cyc}\frac{pq}{(p+r)(q+r)}\geq3$$ or $$\sum_{cyc}r(p-q)^2\geq0$$ and we are done!