Proving $\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc}$ for non-negative $a$, $b$, $c$

Question

I was trying this question with factorization and other similar methods,

Let $a, b, c \geq 0$. Prove that $$\begin{array}{c} \sqrt{a^4+a^2b^2+b^4}+\sqrt{b^4+b^2c^2+c^4}+\sqrt{c^4+c^2a^2+a^4} \\[4pt] \geq a\sqrt{2a^2+bc}+b\sqrt{2b^2+ca}+c\sqrt{2c^2+ab} \end{array}$$

This is one of Hoojoo-Lee's Inequality. This seems very intuitive at first as if we square each term, $$ 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2b^2} \geq 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2bc} \Rightarrow \sum_{cyc}{a^2b^2} \geq \sum_{cyc}{a^2bc} $$ which is quite clear. I noticed it but can not exploit it. May be taking the square on each side could help? But I couldn't find a solution.

Please help!

Answer 1

I think the key tools here are Cauchy-Schwarz and AM-GM. I just bound one term on the LHS. The other two are similar:

$$\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}=\sum_{cyc}\sqrt{(a^4+\frac{a^2b^2}{2})+(b^4+\frac{a^2b^2}{2})}\ge$$$$\frac{1}{\sqrt{2}}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}+\sqrt{b^4+\frac{a^2b^2}{2}}\right)$$$$=\frac{1}{\sqrt{2}}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}+\sqrt{a^4+\frac{a^2c^2}{2}}\right)\ge\sqrt{2}\sum_{cyc}\left(\sqrt{a^4+\frac{a^2b^2}{2}}\sqrt{a^4+\frac{a^2c^2}{2}}\right)^{1/4}$$$$\ge\sqrt{2}\sum_{cyc}\sqrt{a^4+\frac{a^2bc}{2}}=\sum_{cyc}\sqrt{2a^4+a^2bc}$$ I use Cauchy-Schwarz at the first and third inequality, and AM-GM at the second.

Answer 2

By C-S $$\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}=$$ $$=\sqrt{\sum_{cyc}\left(a^4+a^2b^2+b^4+2\sqrt{(a^4+a^2b^2+b^4)(a^4+a^2c^2+c^4)}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}(a^4+a^2b^2+b^4+2(a^4+a^2bc+b^2c^2))}=\sqrt{\sum_{cyc}(4a^4+3a^2b^2+2a^2bc)}.$$ In another hand, by AM-GM $$\sum_{cyc}a\sqrt{2a^2+bc}=\sqrt{\sum_{cyc}(2a^4+a^2bc+2ab\sqrt{(2a^2+bc)(2b^2+ac)})}\leq$$ $$\leq\sqrt{\sum_{cyc}(2a^4+a^2bc+ab(2a^2+bc+2b^2+ac))}=\sqrt{\sum_{cyc}(2a^4+2a^3b+2a^3c+3a^2bc)}.$$ Id est, it's enough to prove that: $$\sum_{cyc}(2a^4-2a^3b-2a^3c+3a^2b^2-a^2bc)\geq0$$ or $$2\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+3\sum_{cyc}(a^2b^2-a^2bc)\geq0,$$ which is true by Schur and Muirhead.

Answer 3

By the Minkowski's inequality, we have \begin{align} \mathrm{LHS} &= \sum_{\mathrm{cyc}}\sqrt{(a^2 + b^2/2)^2 + (b^2\sqrt{3}/2)^2}\\ &\ge \sqrt{\left(\sum_{\mathrm{cyc}} (a^2 + b^2/2)\right)^2 + \left(\sum_{\mathrm{cyc}} b^2\sqrt{3}/2\right)^2 }\\ &= \sqrt{3(a^2+b^2+c^2)^2}. \end{align} Since $x\mapsto \sqrt{x}$ is concave, we have \begin{align} \mathrm{RHS} &= (a+b+c)\sum_{\mathrm{cyc}} \frac{a}{a+b+c}\sqrt{2a^2+bc}\\ &\le (a+b+c)\sqrt{\sum_{\mathrm{cyc}} \frac{a}{a+b+c}(2a^2+bc) }\\ &= \sqrt{a+b+c}\sqrt{2(a^3+b^3+c^3) + 3abc}. \end{align} Thus, it suffices to prove that $$3(a^2+b^2+c^2)^2 \ge (a+b+c)[2(a^3+b^3+c^3) + 3abc]. \tag{1}$$

Let $p=a+b+c, q=ab+bc+ca, r=abc$. (1) is written as $$p^4-6p^2q-9pr+12q^2 \ge 0.$$ Since $q^2 \ge 3pr$, it suffices to prove that $$p^4-6p^2q-3q^2+12q^2 \ge 0$$ that is $$(p^2-3q)^2\ge 0.$$ We are done.