1. ## simple show that

I'm trying to prove that:

(1+x)^-5/2 < 1

Sorry I know it's really easy but I'm just not seeing what to do!

2. The very easiest thing to do, and always do this first, is give a good try at disproving it, rather than assuming it can be shown.

Try x = -0.90 and see what you get for the left-hand side.

3. Hello, hunkydory19!

As given, the statement is not true . . .

Prove: . $(1+x)^{-5/2} \:<\:1$
It is necessary that $x > 0.$

Then we have: . $1 + x \:>\:1$

Raise both sides to the power $\frac{5}{2}\!:\;\;(1 + x)^{\frac{5}{2}} \:>\:1$

Take reciprocals: . $\frac{1}{(1+x)^{\frac{5}{2}}} \:<\:1\quad\Rightarrow\quad(1+x)^{-\frac{5}{2}} \:<\:1$

4. Thank you so much Soboran that was loads of help!

Can I just ask what would happen if it was:

(1+x)^-5/2 > 0

because I wouldn't be able to do the reciprocal stage....

thanks again!

5. Can I just ask what would happen if it was:

(1+x)^-5/2 > 0
$
(1+x)^{-5/2}= ((1+x)^{-5})^{1/2}$

which is greater than or equal to 0 by definition of ^{1/2}