# Show that if $x$ and $y$ are real numbers such that $0<x<1$ and $0<y<1$ [closed]

I'm new to the page, I was hoping that you could help me solve this example so I can get carried away to do my practice exercises.It's an exercise in inequality, maybe it's not difficult, but as soon as I start the topic I have several doubts.

The exercise is: Show that if $x$ and $y$ are real numbers such that $$0<x<1$$ and $$0<y<1,$$ then $$xy(1-x)(1-y)\le \frac{1}{16}$$

Thank you for your help to get a better score on my upcoming test.

## closed as off-topic by GNUSupporter 8964民主女神 地下教會, Did, Severin Schraven, rtybase, Math LoverApr 30 '18 at 22:17

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – GNUSupporter 8964民主女神 地下教會, Did, Severin Schraven, rtybase, Math Lover
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• Can you show us what you've tried and the specific places in which you're feeling doubtful? – Aaron Montgomery Apr 30 '18 at 16:57
• Hint: this is an application of the (a.m.-g.m.-inequality). – GNUSupporter 8964民主女神 地下教會 Apr 30 '18 at 16:58
• @AaronMontgomery I can upload photos to the platform? – user333 Apr 30 '18 at 16:59
• If $x$ and $y$ are near zero, that inequality is clearly false. – Lord Shark the Unknown Apr 30 '18 at 16:59
• It's easy to see the reverse is true $$xy(1-x)(1-y)\le \frac{1}{16}$$ with AM-GM. – GNUSupporter 8964民主女神 地下教會 Apr 30 '18 at 17:02

HINT: Find maximum of the function $f(x)=x(1-x)$.

This requires no knowledge about AM-GM inequality, as in other solutions. Only basic knowledge about quadratic function is needed.

Hint: By AM-GM Inequality, $$x(1-x)y(1-y) \le \left(\frac{x+1-x+y+1-y}{4}\right)^{4}.$$

The inequality should be reversed.

I may show that $x(1-x) \leq 1/4$. By AM-GM, $x + (1-x) \geq 2\sqrt{x(1-x)}$, implying the above.

As $x,y$ are independent, the inequality follows.

Another attempt:

$g(x)=x(1-x)$ ; $f(y)=y(1-y)$.

Note : $x$ and $y$ are independent variables,

$x,y \in (0,1)$.

1) Consider $g'(x)= 1-2x=0$.

Maximum of $g(x)$ is at $x=1/2$.

2)Likewise:

Maximum of $f(y)$ at $y=1/2$.

3)Maximum of $g(x)f(y)$ at $x=y=1/2$:

$\max (f(x)g(y)) =(1/2)^4= (1/16).$

4) Hence $g(x)f(y) \le 1/(16)$, $x,y \in (0,1)$.