Does (a+b)^1/2 equal a^1/2 + b^1/2 - (2ab)^1/2?

If you have ( (a+b)²)^1/2, then because of the binomial theorem, then (a+b)²=a²+b²+2ab.
Inserting this into the equation, then you get (a²+b²+2ab)^1/2, which gives you a+b, since a²+b²+2ab=(a+b)². If you set x=a² and y=b², then you have (x+y+2x^1/2y^1/2)^1/2, which gives you x^1/2+y^1/2. This is also equal to x^1/2 + y^1/2 + (2x^1/2y^1/2)^1/2 - (2x^1/2y^1/2)^1/2, since (2x^1/2y^1/2)^1/2 and -(2x^1/2y^1/2)^1/2 cancel out. What if you removed 2x^1/2y^1/2 from the square root? Then, would you get x^1/2 + y^1/2 - (2x^1/2y^1/2)^1/2?

Does (a+b)^1/2 equal a^1/2 + b^1/2 - (2ab)^1/2?

So, I'm curious why you didn't try this with a few numbers:

 

a=6, b = 9.6 --> sqrt(a+b) = 3.937003937; sqrt(a) + sqrt(b) - sqrt(2ab) = -5.145381508

a=13.3, b = 0.3 --> sqrt(a+b) = 3.687817783; sqrt(a) + sqrt(b) - sqrt(2ab) = 1.369749685

 

and so on...

 

Clearly it doesn't hold except maybe in a few special cases.

Edited December 3, 201312 yr by Bignose

I was just wondering.