Using the AM-GM to find the minimum of $x^2+(4-2x)^2$

Question

I would like to get a min value of the parabola expression $x^2+(4-2x)^2$ with AM-GM inequality. ($x$ is a real number)

$$x^2+(4-2x)^2\geq 2\sqrt{x^2(4-2x)^2 }$$ with equality when $x^2=(4-2x)^2$.

But it gives a wrong value. Why can't we get a correct min value with AM-GM inequality although both $x^2$and $(4-x)^2$ are non negative real numbers?

Answer 1

The AM-GM rule is not for a minimum value of a function. It just says that when $$x^2=(4-2x)^2$$ The two sides becomes equal.

You can test it with $$y=x^2+1^2 \geq 2|x|$$ If you draw the two curves, you will see the minimum point $(0,1)$ is below the $2$ equability points at $x = \pm 1$.

Hope this explained you question.

Answer 2

Let $S=x^2+(4-2x)^2=5x^2-16x+16$. We have $$ 5S=25x^2-80x+80=(25x^2-80x+64)+16=(5x-8)^2+16\geq 16. $$ So $\min(S)=\frac{16}{5}$ which realizes iff $x=\frac{8}{5}$.

Answer 3

Completing squares seems to be the best way for this, AM-GM does not look natural. If you do want to use inequalities, Cauchy-Schwarz is a better bet: $$\left(x^2+(4-2x)^2 \right)(4+1) \ge (2x + \overline{4-2x})^2 = 4^2$$

So $x^2+(4-2x)^2 \ge \frac{16}5$, with equality when $\dfrac{x}2 = 4-2x \implies x = \frac85$.