Optimize $\sqrt{a^2+5bc}+\sqrt{b^2+5ca}+\sqrt{c^2+5ab}\ge \sqrt{a^2+b^2+c^2+17(ab+bc+ca)}.$

Question

Question. Let $a,b,c\ge 0.$ Prove that $$\sqrt{a^2+5bc}+\sqrt{b^2+5ca}+\sqrt{c^2+5ab}\ge \sqrt{a^2+b^2+c^2+17(ab+bc+ca)},$$when $c=\min\{a,b,c\}$ and $(a-b)^2\ge (c-b)(c-a).$

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Here's what I done so far.

By squaring both side, we need to prove $$\sqrt{(a^2+5bc)(b^2+5ca)}+\sqrt{(a^2+5bc)(c^2+5ab)}+\sqrt{(c^2+5ab)(b^2+5ca)}\ge 6(ab+bc+ca). \tag{*}$$ From here, I don't know how to use the hypothesis $c=\min\{a,b,c\}$ and $(a-b)^2\ge (c-b)(c-a).$

I try to verify $(a-b)^2\ge (c-b)(c-a) \iff a^2+b^2-2ab\ge c^2-c(a+b)+ab \iff (a+b)(a+b+c)\ge c^2+5ab.$

Also, we can rewrite as a quadratic of $c$ $$c^2-c(a+b)+3ab-a^2-b^2\le 0.$$

I think if we can divide into two cases.

$\bullet: a\ge b\ge c.$ Let $b=c+t; a=c+t+s$ where $s,t\ge 0.$

$\bullet: b\ge a\ge c.$ Let $a=c+t; b=c+t+s$ where $s,t\ge 0.$

Then, I replace it in $(*)$ but nothing left.

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Hope you can help me. Any ideas and comments is welcome.

Update $1.$

I found that the following inequality is true $$\sqrt{a^2+5bc}+\sqrt{b^2+5ca}+\sqrt{c^2+5ab}\ge \sqrt{a^2+b^2+c^2+\left(4\sqrt{5}+7\right)(ab+bc+ca)}.$$ Equality holds at $a=b>0; c=0.$

Update $2.$

For some related inequalities, see also AOPS

Answer 1

Remark.

The following proof is inspired by Michael Rozenberg's idea. See topic.

Relevant information.

I search for similar problems. See AOPS.

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Proof.

Firstly, we may use $$(a^2+5bc)(b^2+5ca)-\left(ab+\frac{5}{2}c(a+b)\right)^2=5c\left(a+b-\frac{5}{4}c\right)(a-b)^2\ge 0. $$Thus, \begin{aligned} \sqrt{a^2+5bc}+\sqrt{b^2+5ca} \ &=\sqrt{a^2+5bc+b^2+5ca+2\sqrt{(a^2+5bc)(b^2+5ca)}} \\ & \geq \sqrt{a^2+5bc+b^2+5ca+2ab+5ca+5cb} \\ & = \sqrt{(a+b)^2+10c(a+b)}. \end{aligned} It's enough to prove $$\sqrt{(a+b)^2+10c(a+b)}+\sqrt{c^2+5ab} \geq \sqrt{a^2+b^2+c^2+17(ab+bc+ca)}. $$ By squaring both side twice, it's $$ 2\sqrt{\left[(a+b)^2+10c(a+b)\right](c^2+5ab)} \geq 10ab+7(ca+cb),$$ or $$ 4ab(a+b)^2-20a^2b^2+8c^3(a+b)+12abc(a+b)-9c^2(a+b)^2 \geq 0, $$ Now, rewrite as $$ \left[2ab-c(a+b)\right]\left[2(a-b)^2-2ab+ac+cb\right]+2c(a+b)(a+b-2c)^2\geq 0.$$ Use hypothesis $$2(a-b)^2-2ab+ac+cb\ge c(2c-a-b),$$we obtain$$c(a+b-2c)\left[2(a^2+ab+b^2)-3c(a+b)\right]\ge 0.$$The last inequality is obvious since $c$ is the minimal variable.

Hence, the proof is done. Equality holds at $a=b=c.$