For $a,b,c>0;abc=1.$ Prove that $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\ge a+b+c$$ I will post my solution in the answer. Now I'm looking forward to another solution.
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2Some discussion here: https://math.stackexchange.com/questions/3323045/for-positive-a-b-c-with-abc-1-show-that-sum-cyc-sqrta2-a1-geq?noredirect=1 – player3236 Feb 17 '21 at 04:06
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2@player3236 Eww, now I see that my solution is exactly Can's and Mike's ideas. – NKellira Feb 17 '21 at 04:26
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My solution. Assume $(b-1)(c-1)\ge 0 \rightarrow 1=abc\ge (a+b-1)c\rightarrow a+b \le \dfrac{1}{c}+1.$
\begin{align*} \text{LHS}&=\sqrt{\left(a^2+b^2\right)-(a+b)+2+2\sqrt{\left(a^2-a+1\right)\left(b^2-b+1\right)}}+\sqrt{c^2-c+1}\\&\ge \sqrt{t^2-2t+4}+\sqrt{c^2-c+1}=\text{P}\,(\text{where}\,t=a+b)\end{align*} Let $f(t)=P-c-t$ then prove $f(t)\ge 0.$
Since $t-1<\sqrt{(t-1)^2+3}=\sqrt{t^2-2t+4},$ we have $$f'(t)=\dfrac{t-1}{\sqrt{t^2-2t+4}}-1<1-1=0\rightarrow f(t)\ge f\left(\dfrac{1}{c}+1\right)\ge 0.$$ By some simple calculations, we need to prove $3c^2(c-1)^2\ge 0,$ which is true!
Done. Equality holds when $a=b=c=1.$
NKellira
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2You could hide your solution for spoiler prevention (using >! Spoiler text). – David Scholz Feb 17 '21 at 03:57
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2@DavidScholz Could you help me to edit? When I use the hidden option I can't break the line. – NKellira Feb 17 '21 at 04:25
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1Looked good for me, well nevermind. It should work putting ">!" in front of each line. – David Scholz Feb 17 '21 at 04:37