Proving Inequalities for Numbers x and y: Graphs & Algebraic Methods

[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
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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.

 

[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

 

[whkoh]

Thanks for your help!