# Math Help - show that...what method?

1. ## show that...what method?

By considering $\frac{1}{3}x - x^{\frac{1}{3}}$ or otherwise, show that:

$\sqrt[3]{x} < \frac{1}{3}x + \frac{2}{3}$ for all x with 0 < x < 1.

This is another past paper anaylsis question...just wondering if someone could please tell me what method should be used to do this?! I have no idea! Possibly Taylor series?

By considering $\frac{1}{3}x - x^{\frac{1}{3}}$ or otherwise, show that:
$\sqrt[3]{x} < \frac{1}{3}x + \frac{2}{3}$ for all x with 0 < x < 1.
What I would do is multiply both sides of the inequality by 3 and then cube both sides, which maintains the inequality. That gives you $27x < (x + 2)^3$, or $27x < x^3 + 6x^2 + 12x + 8$. Finally, subtracting 27x from both sides gives you $0 < x^3 + 6x^2 -15x + 8$. You can find the roots of the equation $x^3 + 6x^2 - 15x + 8 = 0$ and then analyze the interval $x \in (0, 1)$. Note that 1 is a root of this equation, which should make it easier to factor.