Show that if $x$ and $y$ are real numbers such that $0<x<1$ and $0<y<1$

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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