# Thread: Tricky Inequality

1. ## Tricky Inequality

We got given this problem in a tutorial a few weeks ago and I can't for the life of me figure out how to solve it, probably something simple but would love it if someone could tell me how to go about this.

Prove that:

$\displaystyle \frac{[x^2(3-x)(2+x)]^\frac{1}{3}}{3+x} \leq 1$ if $\displaystyle |x| \leq 1$

Cheers

2. Originally Posted by craig
We got given this problem in a tutorial a few weeks ago and I can't for the life of me figure out how to solve it, probably something simple but would love it if someone could tell me how to go about this.

Prove that:

$\displaystyle \frac{[x^2(3-x)(2+x)]^\frac{1}{3}}{3+x} \leq 1$ if $\displaystyle |x| \leq 1$

Cheers

As $\displaystyle |x|\leq 1$ , we get: $\displaystyle x^2\leq 1\,,\,3-x\leq 4\,,\,2+x\leq 3\,,\,3+x\geq 2$ , so now input this in your expression and you'll get it is less than or equal to $\displaystyle \frac{8^{1/3}}{2}\leq 1$ .

Tonio

3. Ahh thank you, you have no idea how long that was bugging me for!