<s>[INST] 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!!
 [/INST] 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:
$$=\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix}    1&0&0\\
     -1&1& \ \ 0\\
    -2&-2 & \ \ 0   \end{pmatrix}\begin{pmatrix}    1&-1&-2\\
     \ \ 0&\ \ 1&-2\\
    0&0 &0   \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$
$$=\begin{pmatrix}(a-b-2c)&(b-2c)&0\end{pmatrix}\begin{pmatrix}(a-b-2c)\\(b-2c)\\0\end{pmatrix}$$
$$=(a-b-2c)^2+(b-2c)^2$$
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