# Prove $5a^2+b^2+c^2\geq 4ab+2ac$ [closed]

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?

## closed as off-topic by barak manos, Daniel W. Farlow, Brevan Ellefsen, user133281, quid♦Aug 9 '16 at 20:21

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• Muirhead's inequality – Yuriy S Jul 25 '16 at 1:49
• I'm voting to close this question as off-topic because just like every other question of yours, you show no effort whatsoever in attempting to answer it on your own. Asking 'Can anyone help me with it?' is simply not enough. – barak manos Aug 9 '16 at 9:58

Note that $$5a^2+b^2+c^2-4ab-2ac=(2a-b)^2+(a-c)^2\ge 0.$$
$4a^2+b^2+a^2+c^2 \geq 4ab+2ac$ 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.
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 $(1,2,1)$.