# Proving an Equation...

• Sep 11th 2012, 08:35 AM
johnhisenburg87
Proving an Equation...
Did I do the math right? Because the answer doesn't make any sense.

http://i.imgur.com/P97Gs.png
• Sep 11th 2012, 09:17 AM
kalyanram
Re: Proving an Equation...
The math is wrong. Check step 3. How did you get it from step 2?
• Sep 11th 2012, 09:42 AM
johnhisenburg87
Re: Proving an Equation...
I flipped (1/x) and (1/y). I can't do that without flipping the other side, correct?
• Sep 11th 2012, 09:57 AM
johnhisenburg87
Re: Proving an Equation...
So would this be right...?

http://i.imgur.com/XNNhI.png

Where would I go from here?
• Sep 11th 2012, 10:09 AM
MaxJasper
Re: Proving an Equation...
Quote:

Originally Posted by johnhisenburg87
Did I do the math right? Because the answer doesn't make any sense.
http://i.imgur.com/P97Gs.png

$\displaystyle \frac{2 x y}{x+y}\leq \sqrt{x y}$

$\displaystyle \frac{\sqrt{x y}}{x+y}\leq \frac{1}{2}$

$\displaystyle (x+y)^2 \geq 4x y$

$\displaystyle x^2+y^2+2x y \geq 4 x y$

$\displaystyle x^2+y^2-2x y\geq 0$

$\displaystyle (x-y)^2 \geq 0$

Inequality is true if $\displaystyle x \geq y$

Because x & y are interchangeable in the original inequality relation then we also conclude that:

Inequality is true if $\displaystyle y \geq x$

Hence: the only valid solution is x=y

This means inequality is not true for any $\displaystyle x\neq y$, i.e., only euality portion is true for $\displaystyle x=y$

...check this out...

Thanks to Bingk for remarks.
• Sep 11th 2012, 01:58 PM
Bingk
Re: Proving an Equation...
Just some things to note, you're kinda working backwards. You should start with a statement that you knows is true, then manipulate it to get the conclusion.

Try starting with this:

$\displaystyle (\sqrt{x} - \sqrt{y})^2 \geq 0$.

You're allowed to take the square roots because the problem states that x and y are positive real numbers.
Also, note that even if x is less than y, the difference is squared, so the resulting values will always be nonnegative, that's why you know your starting statement is true.

Last thing, on the post above by this by MaxJasper, near the end it says
Quote:

Inequality is true if $\displaystyle x>y$
this is partially true, the inequality is also true if $\displaystyle x<y$ because the difference is squared