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Let $a,b,c>0: abc=1.$ Prove that: $$2(ab+bc+ca)+a+b+c\ge \sqrt{a^2+4(b+c)}+\sqrt{b^2+4(c+a)}+\sqrt{c^2+4(a+b)}$$

I tried to prove the stronger one which is not true: $$2(ab+bc+ca)+a+b +c\ge\sqrt{3(a^2+b^2+c^2)+24(a+b+c)}$$

The inequality seems hard since by Holder: $\sum_{cyc}{\sqrt{a^2+4(b+c)}}=\sum_{cyc}{\sqrt{a}\sqrt{a+\frac{4(b+c)}{a}}}\le\sqrt{(a+b+c)\left(a+b+c+4(a+b+c)(ab+bc+ca)-12\right)}$

The rest is formed p,q,r also without success. I also consider the subsitution: $a=\frac{x}{y}$ but it is quite complicated.

The inequality can be solved by elementory like AM-GM, C-S or high-level method: uvw, BW,.. I have no idea.

Please help me solve problem. Thank you!

Sickness
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    The terms on the RHS strongly suggest quadratic equations may be involved. Since $abc=1$, you might replace $c$ with $\frac 1{ab}$ and see what comes out. – Paul Sinclair Oct 28 '21 at 16:19
  • Thank you, I see. – Sickness Oct 29 '21 at 01:08
  • Can you tell me more? I tried without sucess – Sickness Oct 29 '21 at 12:34
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    Sorry, but I'm just spitballing here. But, if you divide the inequality by $2$ and rearrange a bit, you get: $$ab + bc + ca \ge \frac{-a + \sqrt{a^2+4(b+c)}}2 + \frac{-b + \sqrt{b^2+4(c+a)}}2 + \frac{-c + \sqrt{c^2+4(a+b)}}2$$ The RHS side is the sum of the larger roots of $x^2 + ax -b-c, x^2 + bx - c-a, x^2 + cx- a - b$. I find it hard to believe this is a coincidence. – Paul Sinclair Oct 29 '21 at 13:40
  • I got it. The rest is just C-S. Thank you – Sickness Oct 29 '21 at 13:50
  • Who can think of it? – Sickness Apr 13 '22 at 06:34

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