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Let $ab+bc+ca=1$. Prove that $2 \ge \sqrt{1+a^2} + \sqrt{1+b^2}+\sqrt{1+c^2}-a-b-c \geq \sqrt3 $.

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    $\begingroup$ Jensen , Cauchy - schwarzt , S-O-S , v.v.. $\endgroup$ – Đức Nguyễn Hoàng Jun 23 '17 at 3:07
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The right inequality.

We need to prove that $$ \sum\limits_{\text{cyc}}\sqrt {\left(a+c\right)\left(a+b\right)}\geq a+b+c + \sqrt3\cdot\sqrt {ab+ac+bc}$$ or $$2\left(a+b+c - \sqrt {3(ab+ac+bc)}\right) - \sum_{cyc}\left(\sqrt {a+c} - \sqrt {b+c}\right)^{2}\geq0.$$ But $$ 2\left(a+b+c- \sqrt {3(ab+ac+bc)}\right) - \sum_{cyc}\left(\sqrt {a+c} - \sqrt {b+c}\right)^{2} =$$ $$ = \sum_{cyc}(a+b)^2\left(\frac {1}{a+b+c+ \sqrt {3(ab+ac+bc)}}- \frac {1}{\left(\sqrt {a+c} + \sqrt {b+c}\right)^2}\right)=$$ $$ = \sum_{cyc}\frac {(a-b)^2\left(c + 2\sqrt {(a+c)(b+c)} - \sqrt {3(ab+ac+bc)}\right)}{\left(a+b+c+ \sqrt {3(ab+ac+bc)}\right)\left(\sqrt {a+c} + \sqrt {b+c}\right)^2}=$$ $$ = \sum_{cyc}\frac {(a-b)^2\left(c + \frac {4c^{2} + ab+ac+bc}{2\sqrt {(a+c)(b+c )} + \sqrt {3(ab+ac+bc)}}\right)}{\left(a+b+c + \sqrt {3(ab+ac+bc)}\right)\left(\sqrt {a+c} + \sqrt {b+c}\right)^2}\geq0.$$

The left inequality.

Let $a=\tan\frac{\alpha}{2}$, $b=\tan\frac{\beta}{2}$ and $c=\tan\frac{\gamma}{2}.$

Thus, $\alpha+\beta+\gamma=180^{\circ}$.

Let $\alpha\geq\beta\geq\gamma$.

Thus, $\beta$ and $\gamma$ are acute angles and we need to prove that $$2+\sum\limits_{cyc}f(\alpha)\geq0,$$ where $$f(x)=\tan\frac{x}{2}-\frac{1}{\cos\frac{x}{2}}.$$

But $$f''(x)=\frac{\tan\frac{x}{4}-1}{4\cos^2\frac{x}{4}\left(\tan\frac{x}{4}+1\right)^3}<0$$ for $0<x<\frac{\pi}{2}$, which says that $f$ is a concave function on $\left(0,\frac{\pi}{2}\right)$.

Thus, since $\left(\beta+\gamma,0^{\circ}\right)\succ(\beta,\gamma)$, by Karamata we obtain: $$2+\sum_{cyc}f(\alpha)>2+f(\alpha)+f(\beta+\gamma)+f\left(0^{\circ}\right)=$$ $$=2+f(\alpha)+f(180^{\circ}-\alpha)+f\left(0^{\circ}\right)=1+\tan\frac{\alpha}{2}-\frac{1}{\cos\frac{\alpha}{2}}+\cot\frac{\alpha}{2}-\frac{1}{\sin\frac{\alpha}{2}}=$$ $$=1+\frac{1}{\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}-\frac{1}{\cos\frac{\alpha}{2}}-\frac{1}{\sin\frac{\alpha}{2}}=\frac{\left(1-\sin\frac{\alpha}{2}\right)\left(1-\cos\frac{\alpha}{2}\right)}{\sin\frac{\alpha}{2}\cos\alpha\frac{\alpha}{2}}>0.$$ Done!

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    $\begingroup$ how about the left inequality ? $\endgroup$ – Đức Nguyễn Hoàng Jun 23 '17 at 3:45
  • $\begingroup$ @Đức Nguyễn Hoàng I need to post it. It will take a bit of time. $\endgroup$ – Michael Rozenberg Jun 23 '17 at 3:47
  • $\begingroup$ Okey :) it ok .. I can wait .. $\endgroup$ – Đức Nguyễn Hoàng Jun 23 '17 at 3:50
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    $\begingroup$ There's a small typo in the second to last equation where you wrote a $z$ instead of $c$, but nevertheless this is very beautiful! May I ask how you always manage to come up with these beautiful inequalities? $\endgroup$ – Lazy Lee Jun 23 '17 at 4:43
  • $\begingroup$ @Lazy Lee I fixed my post. Thank you! $\endgroup$ – Michael Rozenberg Jun 23 '17 at 4:52

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