# Proving an inequality of three variables

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!!

• AM-GM Only works with nonnegative real numbers, but you have said that $a,b,c\in \mathbb R$ – Yassir Mar 7 at 10:36
• Note that the expression is quadratic in $a$. So, you need to prove that certain quadratic polynomial is nonnegative for all reals. Can you proceed now? – richrow Mar 7 at 10:51

## 5 Answers

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:

$$=\begin{pmatrix}a&b&c\end{pmatrix}\begin{pmatrix} 1&0&0\\ -1&1& \ \ 0\\ -2&-2 & \ \ 0 \end{pmatrix}\begin{pmatrix} 1&-1&-2\\ \ \ 0&\ \ 1&-2\\ 0&0 &0 \end{pmatrix}\begin{pmatrix}a\\b\\c\end{pmatrix}$$

$$=\begin{pmatrix}(a-b-2c)&(b-2c)&0\end{pmatrix}\begin{pmatrix}(a-b-2c)\\(b-2c)\\0\end{pmatrix}$$

$$=(a-b-2c)^2+(b-2c)^2$$

• Thank you for including your thinking as to how you reached the simplification; most answers to inequality problems just show how to prove the inequality by simplifying it, without justifying how to get to such simplifications. – Aaratrick Mar 8 at 8:19

Hints:

$$\frac12a^2 - 2ab + 2b^2 = \frac12(a-2b)^2 \ge 0$$

$$\frac 12 a^2 - 4ac + 8c^2 = \frac12(a-4c)^2 \ge 0$$

$$a^2 - a(2b+4c) + 2b^2+8c^2 \ge 0$$ since $$\triangle = (2b+4c)^2 -4(2b^2+8c^2) = -4b^2+16bc-16c^2 = -4(b-2c)^2 \le 0$$.

• [+1] Good idea to consider this expression as a quadratic in one of the variables. – Jean Marie Mar 7 at 13:52

$$a^2+2b^2+8c^2=\frac{1}{2}a^2+2b^2+\frac{1}{2}a^2+8c^2$$ 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.

$$a,b,c$$ 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 $$a,b,c$$.