Inequality $a^2+2b^2+8c^2\geq2a(b+2c)$ Can someone prove this inequality for the real numbers a,b,c?
$$a^2+2b^2+8c^2\geq2a(b+2c)$$
I have tried simple manipulation of the terms to get quadratic expressions, but since one cannot factor the $4ac$ on the right side, I abandoned that approach. Then I tried turning it into an expression where I could apply AM-GM, but that did not work either.
Maybe something like Muirhead, though I do not know where to start with that?
Any and all help would be greatly appreciated!!
 A: Your inequation is equivalent to
$$(a-b-2c)^2+(b-2c)^2 \ge 0 \tag{1}$$
The advantage of expression (1) is that it allows to clearly obtain the limit cases where there is an equality sign in (1) instead of a $">"$ symbol, i.e., iff
$$a=2b=4c$$

Edit: One can wonder how I have found expression (1). If you happen to know the concept of matrix associated with a quadratic form, here is the explanation:
$$a^2+2b^2+8c^2-2a(b+2c)$$
$$=\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix}    1&   -1&-2\\
   -1&2&0\\
   -2&0&8 \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$
which can be transformed, using the so-called (incomplete) Cholesky factorization:
$$=\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix}    1&0&0\\
     -1&1& \ \ 0\\
    -2&-2 & \ \ 0   \end{pmatrix}\begin{pmatrix}    1&-1&-2\\
     \ \ 0&\ \ 1&-2\\
    0&0 &0   \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$
$$=\begin{pmatrix}(a-b-2c)&(b-2c)&0\end{pmatrix}\begin{pmatrix}(a-b-2c)\\(b-2c)\\0\end{pmatrix}$$
$$=(a-b-2c)^2+(b-2c)^2$$
A: Hints:
$$\frac12a^2 - 2ab + 2b^2 = \frac12(a-2b)^2 \ge 0$$
$$\frac 12 a^2 - 4ac + 8c^2 = \frac12(a-4c)^2 \ge 0$$
A: $a^2 - a(2b+4c) + 2b^2+8c^2 \ge 0$ since $\triangle = (2b+4c)^2 -4(2b^2+8c^2) = -4b^2+16bc-16c^2 = -4(b-2c)^2 \le 0$.
A: $$a^2+2b^2+8c^2=\frac{1}{2}a^2+2b^2+\frac{1}{2}a^2+8c^2$$
Now By AM-GM :
$$\cases{\frac{1}{2}a^2+ 2b^2 \geq 2ab \\ \frac{1}{2}a^2+8c^2\geq4\sqrt{a^2c^2}=4ac}$$
Adding them together:
$$a^2+2b^2+8c^2 \geq2ab+4ac$$
And that’s a good place to stop.
A: $a,b,c$ are real numbers.
$$a^2+2b^2+8c^2\geq2a(b+2c)~~~~(1)$$
if for all real values of $a$
$$a^2-a(b+2c)^2+3b^2+8c^2\ge 0~~~~~(2)$$
$$\implies B^2 \le 4AC$$, as $A>0$, then
$$(b+2c)^2 \le 4(3b^2+8c^2) \implies -11b^2+4bc-28c^2\le 0$$
$$\implies 11b^2-4bc+28c^2\ge 0~~~~~(3)$$
In this quadratic $$D=B^2-4AC=4c^2-11.28 c^2 \le 0~~~~(4)$$
So (3) holds for all real values of $b$ and  (4) holds for all real values of $c$. Hence, this proves that (1), holds for all real values of $a,b,c$.
