How to prove this inequality $\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}\;?$ 
Let $a,b,c\ge 0$, and assume $a+b+c=3$.
I'd like to show that
$$\frac{a}{\sqrt{b^2+b+1}}+\frac{b}{\sqrt{c^2+c+1}}+\frac{c}{\sqrt{a^2+a+1}}\ge\sqrt{3}$$

My try: Use Hölder inequality
$$\left(\sum_{cyc}\dfrac{a}{\sqrt{b^2+b+1}}\right)\left(a\sqrt{b^2+b+1}\right)^2\ge (a+b+c)^3\,,$$
then we only need to prove this
$$\sum_{cyc}a(b^2+b+1)=\sum_{cyc}(ab^2+ab+a)\le 9$$
but
$$ab^2+bc^2+ca^2+ab+bc+ac\le 6$$
is not true.
So I can't prove it.
 A: I'm going to use two facts. The first one can be obtained as following:
$$\begin{align}
ab + bc + ca &\leqslant a^2 + b^2 + c^2, \\
3(ab + bc + ca) &\leqslant (a + b + c)^2 = 9, \\
ab + bc + ca &\leqslant 3.
\end{align}$$
And the second one:
$$\frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}}$$
(here's Wolfram Alpha visualization, a bit later I'm going to prove it more rigorously).
So, putting it all together:
$$
\begin{align}
\frac{a}{\sqrt{b^2 + b + 1}} + \frac{b}{\sqrt{c^2 + c + 1}} + \frac{c}{\sqrt{a^2 + a + 1}} &\geqslant \frac{\sqrt{3}}{2}(a + b + c) - \frac{ab + bc + ca}{2\sqrt{3}} \\
&\geqslant \frac{3\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = \sqrt{3}
\end{align}$$

Lemma. $\displaystyle \frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}},\ \forall x \geqslant 0$.
Note that $\displaystyle \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}} = \frac{1}{2\sqrt{3}}(3 - x) < 0$ for $x > 3$. So, for these $x$ we get obvious inequality that square root of something is greater than $0$.
For $x \in [0, 3]$:
$$
\begin{align}
\frac{1}{\sqrt{x^2 + x + 1}} \geqslant \frac{\sqrt{3}}{2} - \frac{x}{2\sqrt{3}} &\Longleftrightarrow 12 \geqslant (x^2 + x + 1)(3 - x)^2 \\
&\Longleftrightarrow x^4 - 5x^3 + 4x^2 +3x - 3 \leqslant 0 \\
&\Longleftrightarrow (x - 1)(x^3 - 4x^2 + 3) \leqslant 0 \\
&\Longleftrightarrow (x - 1)^2(x^2 - 3x - 3) \leqslant 0
\end{align}$$
which is true because roots of $x^2 - 3x - 3$ are:
$x_1 = \frac{1}{2}(3 - \sqrt{21}) < 0$,
$x_2 = \frac{1}{2}(3 + \sqrt{21}) > 3$.
So for $x \in [0, 3]$ holds $x^2 - 3x - 3 < 0.$
