# Simple proof (hopefully) when $0 <x<1$

How can one prove that $x(2-x)^2 \leq 1$ when $0<x<1$ ? Alternatively, of course, this is $4x-4x^2+x^3 -1 \leq 0$ but I don't know where to go from there. Is it enough to show the inequality holds for both bounds of $x$, i.e. $4(0) - 4(0)^2 + 0^3 -1 = -1 < 0$ and $4(1) -4(1)^2 + 1^3 -1 = 0 \leq 0$ ? How do I show it's always true?

Thank you!

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Thanks to all! I apologize for posing a false statement, and appreciate all the answers both pointing this out, and providing general techniques. –  Angada Sep 14 '11 at 22:02

For a continuous function, you can check both ends, any point where the derivative is zero, and any point the derivative fails to exist. The range of the function will be from the minimum to the maximum of those.

You can look at the extreme value theorem

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The claim is false. For example, when $x=0.5$, $x(2-x)^2=1.125$.

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The claim is wrong as Chris points out.

But I will show how to compute the maximum value of this function in that interval using elementary inequalities. $$x(2-x)^2 = \frac{1}{2} (2x) \cdot (2-x) \cdot (2-x) \leq \frac{1}{2} \left( \frac{2x+(2-x)+(2-x)}{3} \right)^3 = \frac{1}{2} \cdot \left(\frac{4}{3} \right)^3 = \frac{32}{27}.$$ where the inequality follows from the AM-GM inequality. (Convince yourselves that all the terms are nonnegative, so AM-GM can be applied.) Equality occurs when $2x = 2-x = 2-x$, i.e., when $x = 2/3$. The maximum value therefore is $32/27$.

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If $f(x) = x(2-x)^2$, then $f'(x) = (2-x)^2 - 2x(2-x) = (2-x)(2-3x)$. The critical points are at $x=2$ and $x=\frac{2}{3}$. On $(-\infty,\frac{2}{3})$, $f'(x)\gt 0$ so the function is increasing; on $(\frac{2}{3},2)$, $f'(x)\lt 0$ so the function is decreasing; and on $(2,\infty)$, $f'(x)\gt 0$, so the function is increasing.

On $[0,1]$ then the maximum is achieved at $\frac{2}{3}$, where it is equal to $\frac{32}{27}\gt 1$.

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