Prove this inequality $2(a+b+c)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$ For $a,b,c$ are positive real numbers satisfy $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$2\left(a+b+c\right)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$$

We have:$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\Leftrightarrow \:\left(a+b+c\right)^2\ge \:9\Leftrightarrow \:a+b+c\ge \:3$
By Cauchy-Schwarz: $R.H.S^2\le \left(1+1+1\right)\left(a^2+b^2+c^2+27\right)$
$=3\left(a^2+b^2+c^2+9\right)\Rightarrow R.H.S\le \sqrt{3\left(a^2+b^2+c^2\right)+27}$
Need to prove $4(a+b+2)^2\ge 3(a^2+b^2+c^2)+27$
$\Leftrightarrow \left(a+b+c\right)^2+3\left(a+b+c\right)^2\ge 3\left(a^2+b^2+c^2\right)+27$
$\Leftrightarrow \left(a+b+c\right)^2\ge 3\left(a^2+b^2+c^2\right)$ It's wrong. Help me
 A: Since the function $f(t) := \sqrt{1+t}$ is concave in $[-1, +\infty)$, we have that
$$
f(t) \leq f(3) + f'(3) (t-3) 
\qquad \forall t\geq -1,
$$
i.e.
$$
f(t) \leq 2 + \frac{1}{4}(t-3) = \frac{5}{4}  + \frac{1}{4} t
\qquad \forall t\geq -1.
$$
Using this inequality we have that
$$
\sqrt{a^2+3} = a \sqrt{1+ 3/a^2} \leq
a \left[\frac{5}{4} + \frac{1}{4}\cdot\frac{3}{a^2}\right]
= \frac{5}{4} a + \frac{3}{4}\cdot \frac{1}{a},
$$
and a similar inequality holds also for $\sqrt{b^2+3}$ and $\sqrt{c^2+3}$.
Finally, using the condition $a+b+c = \frac{1}{a}
+ \frac{1}{b} + \frac{1}{c}$,
$$
\sqrt{a^2+3} + \sqrt{b^2+3} + \sqrt{c^2+3} 
\leq 
\frac{5}{4} (a + b + c) + \frac{3}{4}\left(\frac{1}{a}
+ \frac{1}{b} + \frac{1}{c}\right) 
= 2(a+b+c).
$$ 
A: An alternative proof to that presented by @Rigel and one that does not require any results from calculus/differentiation is as follows:
Let $\mathbb{u}\equiv\left(\sqrt{a},\sqrt{b},\sqrt{c}\right)$ and $\mathbb{v}\equiv\left(\sqrt{a+\frac{3}{a}},\sqrt{b+\frac{3}{b}},\sqrt{c+\frac{3}{c}}\right)$. From these definitions one has:
$$ |\mathbb{u}\cdot\mathbb{v}|=\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3},\qquad \mbox{and} $$
$$ ||\mathbb{u}||\!\cdot\!||\mathbb{v}||=\sqrt{a+b+c}\sqrt{a+b+c+3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}=\sqrt{a+b+c}\sqrt{4\left(a+b+c\right)}=2(a+b+c). $$
The result follows by application of the Cauchy-Schwarz Inequality.
Using the same method of proof, one can obtain the following more general inequality:
$$ \sqrt{K+1}(a+b+c)\geq\sqrt{a^2+K}+\sqrt{b^2+K}+\sqrt{c^2+K}, $$
for any positive $K$.
A: We need to prove that
$$\sum_{cyc}\left(2a-\sqrt{a^2+3}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-1}{2a+\sqrt{a^2+3}}\geq0$$ or 
$$\sum_{cyc}\left(\frac{a^2-1}{2a+\sqrt{a^2+3}}+\frac{1}{4}\left(\frac{1}{a}-a\right)\right)\geq0$$ or
$$\sum_{cyc}(a^2-1)\left(\frac{1}{2a+\sqrt{a^2+3}}-\frac{1}{4a}\right)\geq0$$ 0r
$$\sum_{cyc}\frac{(a^2-1)(2a-\sqrt{a^2+3})}{a(2a+\sqrt{a^2+3})}\geq0$$ or
$$\sum_{cyc}\frac{(a^2-1)^2}{a(2a+\sqrt{a^2+3})^2}\geq0.$$
Done!
