# Inequality- absolute value function

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• February 18th 2013, 03:44 AM
sachinrajsharma
Inequality- absolute value function
Prove that if the numbers x,y are of one sign, then $|\frac{x+y}{2}-\sqrt{xy}|+|\frac{x+y}{2}+\sqrt{xy}|=|x|+|y|$
• February 18th 2013, 03:57 AM
Plato
Re: Inequality- absolute value function
Quote:

Originally Posted by sachinrajsharma
Prove that if the numbers x,y are of one sign, then $|\frac{x+y}{2}-\sqrt{xy}|+|\frac{x+y}{2}+\sqrt{xy}|=|x|+|y|$

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• February 18th 2013, 08:49 PM
sachinrajsharma
Re: Inequality- absolute value function
I am sorry as I am new to this forum I didnt know that I have to show my work as well.. moreover this is not my homework what I have done

$|\frac{x+y}{2}-\sqrt{xy}|+|\frac{x+y}{2}+\sqrt{xy}|=|x|+|y|$

Solving R.HS. i.e. $|\frac{x+y}{2}-\sqrt{xy}|= |\frac{x+y -\sqrt{xy}}{2}| = \frac{(\sqrt{x}-\sqrt{y})^2}{2}$

$|\frac{x+y}{2}+\sqrt{xy}| = |\frac{ (\sqrt{x}+\sqrt{y})^2}{2} |$
Please suggest further..
• February 19th 2013, 02:41 AM
ibdutt
Re: Inequality- absolute value function
Firstly x and y have to be of same sign for otherwise the we will have square root of a negative number which is not real. Thereafter we know that (a+b)^2+(a-b)^2= 2(a^2+b^2)