How can we prove this inequality? Let $a$ , $b$ and $c$ be positive real numbers and $a+b+c=1$ How can we show this inequality?
$$a^2+b^2+c^2+2\sqrt{3abc}\le 1$$
Thanks.
 A: We have that
$$1=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca),$$
so it suffices to show that
$$ab+bc+ca\geq \sqrt{3abc}.$$
Now $X^2+Y^2+Z^2\geq XY+YZ+XZ$ implies
$$ab+bc+ca=(\sqrt{ab})^2+(\sqrt{bc})^2+(\sqrt{ca})^2
\\\geq b\sqrt{ac}+c\sqrt{ab}+a\sqrt{bc}
=(\sqrt{a}+\sqrt{b}+\sqrt{c})\sqrt{abc}.$$
Hence, we still have to prove that
$$\sqrt{a}+\sqrt{b}+\sqrt{c}\geq \sqrt{3}$$
which holds because $\sqrt{x}$ is concave and
$$\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{3}\geq \sqrt{\frac{a+b+c}{3}}=\frac{1}{\sqrt{3}}.$$
A: Set $a=x^2$ , $b=y^2$ and $c=z^2$ and define
$$f(x,y,z)=x^4+y^4+z^4+2\sqrt{3}xyz-1$$
and 
$$g(x,y,z)=x^2+y^2+z^2-1=0$$
By application of Lagrange method, we have
$$\nabla f(x,y,z)=\lambda \nabla g(x,y,z)$$
thus
\begin{cases}
4x^3+2\sqrt{3}yz=2\lambda x\\
4y^3+2\sqrt{3}xz=2\lambda y\\
4z^3+2\sqrt{3}xy=2\lambda z\\
\end{cases}
so
\begin{cases}
2x^4+\sqrt{3}xyz=\lambda x^2\\
2y^4+2\sqrt{3}xyz=\lambda y^2\\
2z^3+2\sqrt{3}xyz=\lambda z^2\\
\end{cases}
it is equivalent by 
\begin{cases}
2x^4-2y^4=\lambda (x^2-y^2)\\
2y^4-2z^4=\lambda (y^2-z^2)\\
2z^4-2x^4=\lambda (z^2-x^2)\\
\end{cases}
or
\begin{cases}
x^2+y^2=\frac 12\lambda \\
y^2+z^2=\frac 12\lambda\\
z^2+x^2=\frac 12\lambda \\
\end{cases}
indeed
$$x^2+y^2=y^2+z^2=z^2+x^2$$
as a result
$$x^2=y^2=z^2\tag 1$$
on the other hand
$$x^2+y^2+z^2-1=0\tag 2$$
$(1)$ and $(2)$
$$x=y=z=\frac{1}{\sqrt{3}}\tag 3$$
then 
$$f(x,y,z)\le f\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)=0$$
finally
$$x^4+y^4+z^4+2\sqrt{3}xyz-1\le 0$$
or
$$a^2+b^2+c^2+2\sqrt{3abc}\le 1$$
