# More inequalities

1. Jan 7, 2006

### whkoh

The numbers x and y satisfy $0 < x \leq a^2, 0 < y \leq a^2, xy \geq a^2$ where $a \geq 1$.

By sketching suitable graphs or otherwise, show that
$$x + y \geq 2a$$ and $x \leq a^{2}y \leq a^{4}x$
---
I don't know what to sketch (tried $x \leq 1, y \leq 1, xy \leq 1$), so I tried algebraic methods.

For the 1st:
$$y < x+y$$
$$y^2 < x^2 + 2xy + y^2$$
$$x^2 + 2xy > 0$$

For the second one:
$$x \leq a^2 \leq xy$$
$$a^2 \leq xy \leq a^{2}y$$

I'm really lost on this.

Last edited: Jan 7, 2006
2. Jan 7, 2006

### marlon

For the first one, i get this

$$xy \leq a^2y$$
$$xy \leq a^2x$$
$$2xy \leq a^2(x+y)$$
$$2a^2 \leq 2xy \leq a^2(x+y)$$

For the second one
$$x \leq a^2$$
$$xy \leq a^2y$$
$$a^2 \leq xy \leq a^2y$$
$$x \leq a^2 \leq xy \leq a^2y$$
$$x^2 \leq a^2y$$

$$y \leq a^2$$
$$a^2y \leq a^4$$ since y has maximum value equal to a²
$$a^2y \leq a^4 \leq a^2xy \leq a^4x$$
$$a^2y \leq a^4x$$

I know it is not exactly what they are asking, but it is as far as i can get from the top of my head. Others will most certainly correct me...

marlon

Last edited: Jan 7, 2006
3. Jan 7, 2006

### whkoh

Thanks for your help!