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Let $a,b,c>0: \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}.$ Prove that: $$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+3\ge\sqrt{a^2+b^2+1}+\sqrt{b^2+c^2+1}+\sqrt{c^2+a^2+1}$$

My attempts: I thought that AM-GM may be able to solve.

By AM-GM inequality: $$\frac{ab}{c}+\frac{bc}{a}\ge2b; \frac{bc}{a}+\frac{ca}{b}\ge2c;\frac{ca}{b}+\frac{ab}{c}\ge2a$$ Thus, It is desired to prove the following one: $$a+b+c+3\ge\sum_{cyc}{\sqrt{a^2+b^2+1}}$$ Also by condition: $a+b+c\ge6$, the rest is not true by $\sum_{cyc}{\sqrt{a^2+b^2+1}}\ge9$.

I also tried strong result: $$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge\sqrt{3(a^2+b^2+c^2)}$$ without success. Whether uvw method work there?

If anyone find something useful to get proof for the problem, please tell me. Thank you for your time and your knowledge!

Sickness
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    Unfortunately you cannot prove it this way. Take for example $a = 2, b = 3, c = 1.5$ which also fails the condition $a + b + c ≥ 6$. – Toby Mak Oct 23 '21 at 07:07
  • Yes, it seems I am lost – Sickness Oct 23 '21 at 07:15
  • I trying: $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+3\ge\sqrt{3(2a^2+2b^2+2c^2+3)}$ Anyone can help me the rest? – Sickness Oct 23 '21 at 13:04
  • @Mars If you were able to do it using uvw, why don't you add a partial answer? I mean, what I've tried is to put $x= \frac 1a, y=\frac 1b, z =\frac 1c$ and so on and I cleared the denominators. This led to some kind of in-homogenous inequality though, so that concerns me. I think the idea here, is to try and guess a "general homogenous inequality" of which this particular inequality is a special case with $x+y+z = \frac 32$. To me, it looks like the general form yields to Cauchy-Schwarz, because the RHS of the $xyz$ expression continues to look like a summation of square roots. – Sarvesh Ravichandran Iyer Oct 25 '21 at 11:56
  • @Teresa Lisbon Thank you for your sharing. Actually, I just guess uvw helps and I am not good at using this method. – Sickness Oct 25 '21 at 12:36
  • Whether we get a stronger lemma than one I tried? – Sickness Oct 25 '21 at 12:37
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    @Mars I see! No problem, I will try to plough through what's here. I am not particularly strong with inequalities, but I try to make out what I can from similar questions, because I don't think it's as ad-hoc as we often make it out to be. I've seen some similar questions, so I feel I can have a reasonable crack at this one. – Sarvesh Ravichandran Iyer Oct 25 '21 at 12:48
  • There is a nice and short proof by AM-GM. If no one post solution, I will. I wonder how author can come up with such nice and hard inequality like this one. I am very curious! The only trouble is time. Also please show more what you got dear friend, I keep mind mine because I am limited post. – Sickness Oct 25 '21 at 13:57
  • Please share me. Thank you – Sickness Oct 25 '21 at 14:59
  • @3017 Why you don't give proof? Please help me – Sickness Oct 26 '21 at 02:19
  • No one help me ? – Sickness Oct 27 '21 at 02:57
  • Maybe squaring both side – Sickness Oct 28 '21 at 10:22

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