Proof.... Struggling for some reason?

Question

For a>0 and b>0 with a not equal to b, show that the square root of (ab) < (a+b)/2

I feel like I should be solving this by contradiction, and I keep trying, but I haven't managed to come up with it yet. I can't get it to work for some reason.

Answer 1

I think by contradiction works:
if 2sqrt(ab)> or = a +b then
4ab >= (a + b)^2 = a^2 + 2ab + b^2 or
0 >= a^2 - 2ab +b^2 = (a - b)^2 which is strictly >0 since a is not equal to b.
The only question I had (for myself) is squaring both sides of the inequality. Since everything is posit. and the squaring function for positive values is monotonically increasing, it works. Questions: [e-mail]

Answer 2

Start with sqrt(ab) < (a+b)/2

Square both sides

ab < 1/4*(a^2+b^2+2ab) subtract 4ab from both sides

0 < a^2- 2ab + b^2 = (a - b)^2 ---> 0 < (a-b)^2

But that implies a-b is an imaginary number hence proving the relationship.

Answer 3

Let's manipulate your inequaltity a bit:
We have
2√ab < a+b.
Squaring both sides,
4ab < (a+b)² = a² + 2ab + b²
2ab < a² + b²
0 < a² - 2ab + b² = (a-b)²,
which is true, since a <> b.
Working this backwards, yields your result!