Prove $5a^2+b^2+c^2\geq 4ab+2ac$ I saw this question recently:

Let $a,b,c$ be real numbers. Prove $5a^2+b^2+c^2\geq 4ab+2ac$.

I feel like this is something with AM-GM inequality. Can someone help me with it?
 A: Notice that:
$4a^2+b^2+a^2+c^2 \geq 4ab+2ac$ By AM-GM inequality.
EDIT:
The reason why I come up with such idea is because $b$ and $c$ in RHS is independent. So $ab$ comes from one AM-GM and $ac$ comes from the other. 
Notice that $4ab = 2\sqrt{4a^2b^2} $ since the $2$ is the constant deriving from AM-GM, one shall find the suitable combination of coefficient of $a^2$ and $b^2$ in LHS, $4 = 2^2$ , one combination is $2a^2$,$2b^2$, but the coefficient of $b^2$ in LHS is 1. So it is in vain, then I try $4a^2$,$b^2$. That works.    
A: The quadratic form $5a^2-4ab+b^2-2ac+c^2$ is associated with the symmetric matrix
$$ M=\begin{pmatrix}5 & -2 & -1 \\ -2 & 1 & 0 \\ -1 & 0 & 1\end{pmatrix} $$
that is a positive semi-definite matrix by Sylvester's criterion:
$$ 5>0,\qquad \det\begin{pmatrix} 5 & -2 \\ -2 & 1 \end{pmatrix}>0,\qquad \det(M)=0.$$
It follows that for every $(a,b,c)\in\mathbb{R}^3$,
$$  5a^2-4ab+b^2-2ac+c^2 \geq 0 $$
as wanted, and equality is achieved only by $(a,b,c)=\lambda(1,2,1)$, since $\ker(M)$ is generated by $(1,2,1)$.
A: Note that 
$$5a^2+b^2+c^2-4ab-2ac=(2a-b)^2+(a-c)^2\ge 0.$$
