Hello,

I'm having trouble with this maths question and was wondering if someone could help me?? I've been asked to use the Mean Value Theorem to show that:

sqrt (x*y) < 1/2 * (x + y) if 0 < x < y

And we were hinted to use the function f(x) = sqrt x

2. Given the interval [x,y] for the function $f(z) = \sqrt{z}$ (so I don't confuse the variable of the function with the endpoint of my interval), there exists some c such that:
$f'(c) = \frac{f(y) - f(x)}{y-x} = \frac{\sqrt{y} - \sqrt{x}}{y - x}$

Since $0 < x < y$, we know that f'(c) > 0.

Now, the trick is to multiply top and bottom by "1", that is: $\sqrt{y} - \sqrt{x}$. So we get:

$f'(c) = \frac{\sqrt{y} - \sqrt{x}}{y-x} \cdot \left(\frac{\sqrt{y} - \sqrt{x}}{\sqrt{y} - \sqrt{x}}\right) > 0$

$f'(c) = \frac{y - 2\sqrt{xy} + x}{\mbox{(Some denominator)}} > 0$

Multiply both sides by the denominator and you'll (hopefully) reach your inequality.

3. Thankyou so much!! You really helped me!!