Prove that $\sum\limits_{cyc}{\frac{1}{2(a-b)^2+ab}} \geq \frac{16(ab+bc+ca)^3(abc-1)}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+3$

Question

PROBLEM.Let $ a,b,c > 0$ such that $a+b+c=3$. Prove that $$\sum_{cyc}{\frac{1}{2(a-b)^2+ab}} \geq \frac{16(ab+bc+ca)^3(abc-1)}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+3.$$

By AM-GM : $$\sum_{cyc}\frac{1}{2(a^2+b^2)-3ab} +16(a+b+c) \geq 42abc +9$$ Proof $$42abc-16(a+b+c)+6 \geq \frac{16(ab+bc+ca)^3(abc-1)}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\tag{$*$}$$ Thus, it's enough to prove that: $$\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab+bc+ca$$

Therefore $(*)$ is true by AM-GM. And we are done!

What other solutions are there?

Answer 1

Your first step leads to a wrong inequality.

It's enough to prove that: $$\frac{16(1-abc)(ab+ac+bc)^3}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\geq42(1-abc)$$ and since by AM-GM $1-abc\geq0,$ we need to prove that: $$16(ab+ac+bc)^3\geq42(\sqrt{a}+\sqrt{b}+\sqrt{c}),$$ which is wrong for $b=c\rightarrow0^+$.

By the way, by C-S $$\sum_{cyc}\frac{1}{2a^2-3ab+2b^2}=\sum_{cyc}\frac{c^2}{c^2(2a^2-3ab+2b^2)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(4a^2b^2-3a^2bc)}$$ and since by AM-GM $$\sum_{cyc}\sqrt{a}\leq\sum_{cyc}\frac{a+1}{2}=3,$$ it's enough to prove that: $$\frac{9}{\sum\limits_{cyc}(4a^2b^2-3a^2bc)}+\frac{16(ab+ac+bc)^3(1-abc)}{3}\geq3,$$ which is true, but my proof of this statement is not nice.

I hope, it will help.

Answer 2

By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\sum_{\mathrm{cyc}}{\frac{1}{2(a-b)^2+ab}} \ge \frac{(c + a + b)^2}{\sum_{\mathrm{cyc}} c^2(2a^2 + 2b^2 - 3ab)}. \tag{1}$$

By AM-GM, we have $abc \le (a + b + c)^3/27 = 1$ and $\sqrt{a} + \sqrt b + \sqrt c \le \frac{a + 1}{2} + \frac{b+1}{2} + \frac{c+1}{2} = 3$.

It suffices to prove that $$\frac{9}{\sum_{\mathrm{cyc}} c^2(2a^2 + 2b^2 - 3ab)} + \frac{16(ab + bc + ca)^3(1 - abc)}{3} \ge 3. \tag{2}$$

We use pqr method.

Let $p = a + b + c = 3, q = ab + bc + ca, r = abc$.

(2) is written as $$\frac{9}{4q^2 - 33r} + \frac{16q^3(1 - r)}{3} \ge 3. \tag{3}$$

Using $pq \ge 9r$ (by AM-GM), it suffices to prove that $$\frac{9}{4q^2 - 33r} + \frac{16q^3(1 - q/3)}{3} \ge 3. \tag{4}$$

We split into two cases.

Case 1: $q > 9/4$

Using degree three Schur, we have $p^3 - 4pq + 9r \ge 0$ which results in $r \ge 4q/3 - 3$.

It suffices to prove that $$\frac{9}{4q^2 - 33\cdot (4q/3 - 3)} + \frac{16q^3(1 - q/3)}{3} \ge 3$$ which is true.

Case 2: $q \le 9/4$

It suffices to prove that $$\frac{9}{4q^2} + \frac{16q^3(1 - q/3)}{3} \ge 3$$ which is true.

We are done.