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I'm looking for some ideas to prove the following inequality.

Problem. Let $a,b,c$ be three non-negative real numbers with $a+b+c=4.$ Prove that$$\sqrt{a+5bc+4}+\sqrt{b+5ca+4}+\sqrt{c+5ab+4}\ge 2\sqrt{6}\sqrt{ab+bc+ca}.$$When does equality hold?
Source: Tran Ngoc Khuong Trang@IIBAAK.

Equality holds iff $a=b=c=\dfrac{4}{3}$ or $a=b=2,\ c=0$ and its permutations.

Here is an idea.

By a homogenization, we'll prove the inequality$$\color{black}{\sum_{\mathrm{cyc}}\sqrt{a(a+b+c)+20bc+(a+b+c)^2}\ge 4\sqrt{6}\sqrt{ab+bc+ca}}$$or$$\color{black}{\sum_{\mathrm{cyc}}\sqrt{2a^2+b^2+c^2+3ab+3ac+22bc}\ge 4\sqrt{6}\sqrt{ab+bc+ca}.}$$Now, we may use Holder inequality as follow$$\color{black}{\left(\sum_{\mathrm{cyc}}\sqrt{2a^2+b^2+c^2+3ab+3ac+22bc}\right)^2\cdot\sum_{\mathrm{cyc}}\frac{(a^2+b^2+c^2+2ab+2ac+6bc)^3}{2a^2+b^2+c^2+3ab+3ac+22bc}\ge }$$$$\ge \left[\sum_{\mathrm{cyc}}(a^2+b^2+c^2+2ab+2ac+6bc)\right]^3$$and it remains to prove$$\color{black}{\left[4ab+4bc+4ca+3(a+b+c)^2\right]^3\ge 96(ab+bc+ca)\cdot\sum_{\mathrm{cyc}}\frac{(a^2+b^2+c^2+2ab+2ac+6bc)^3}{2a^2+b^2+c^2+3ab+3ac+22bc}.}$$This inequality is true and has an ugly equivalent SOS expression.

Also, I wonder that Mixing Variables technique helps here (MV method). Could someone please answer my question?

Hope you share some ideas to solve this inequality. Thank you for your interest!

Michael Rozenberg
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1 Answers1

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Yes, Mixing Variables technique helps.

By your work we need to prove that: $$\sum_{cyc}\sqrt{2a^2+3ab+3ac+b^2+c^2+22bc}\geq4\sqrt{6(ab+ac+bc)}.$$ Now, let $a\geq b\geq c$.

We'll prove that $$\sqrt{2a^2+3ab+3ac+b^2+c^2+22bc}+\sqrt{2b^2+3ab+3bc+a^2+c^2+22ac}\geq$$ $$\geq\sqrt{3a^2+3b^2+18ab+50ac+50bc+4c^2}.$$

Indeed, after squaring of the both sides we need to prove that: $$2\sqrt{(2a^2+3ab+3ac+b^2+c^2+22bc)(2b^2+3ab+3bc+a^2+c^2+22ac)}\geq12ab+25ac+25bc+2c^2$$ or $$(a-b)^2(8a^2+8b^2+52ab+188ac+188bc-349c^2)\geq0,$$ which is true for $a\geq b\geq c$.

Thus, it's enough to prove that: $$\sqrt{3a^2+3b^2+18ab+50ac+50bc+4c^2}+\sqrt{2c^2+3ac+3bc+a^2+b^2+22ab}\geq4\sqrt{6(ab+ac+bc)}$$ or $$2\sqrt{(3a^2+3b^2+18ab+50ac+50bc+4c^2)(2c^2+3ac+3bc+a^2+b^2+22ab)}+$$ $$+4a^2+4b^2-56ab-43ac-43bc+6c^2\geq0$$ and since $4a^2+4b^2\geq8ab,$ it's enough to prove: $$2\sqrt{(3a^2+3b^2+18ab+50ac+50bc+4c^2)(2c^2+3ac+3bc+a^2+b^2+22ab)}\geq$$ $$\geq48ab+43ac+43bc-6c^2$$ or $f(a)\geq0,$ where $$f(a)=12a^4+336a^3b-696a^2b^2+336ab^3+12b^4+$$ $$+(236a^3+724a^2b+724ab^2+236b^3)c-(1209a^2+1496ab+1209b^2)c^2+$$ $$+964(a+b)c^3-4c^4.$$ But, $$f'(a)=48a^3+1008a^2b-1392ab^2+336b^3+$$ $$+(708a^2+1448ab+724b^2)c-(2418a+1426b)c^2+964c^3$$ and $$f''(a)=144a^2+2016ab-1392b^2+1416ac+1448bc-2418c^2\geq0,$$ which says $$f'(a)\geq48b^3+1008b^3-1392b^3+336b^3+$$ $$+(708b^2+1448b^2+724b^2)c-(2418b+1426b)c^2+964c^3=$$ $$=c(b-c)(2880b-964c)\geq0$$ and we obtain: $$f(a)\geq12b^4+336b^4-696b^4+336b^4+12b^4+$$ $$+(236b^3+724b^3+724b^3+236b^3)c-(1209b^2+1496b^2+1209b^2)c^2+$$ $$+964(b+b)c^3-4c^4=4c(b-c)^2(480b-c)\geq0,$$ which finishes a proof.

Michael Rozenberg
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