If $1<a<b$, which of the following is larger than the other: $a\sqrt[3]{b^2}$ and $b\sqrt[3]{a^2}$?

Question

Given that $1<a<b$, how can you determine which is the larger, out of $a\sqrt[3]{b^2}$ and $b\sqrt[3]{a^2}$?

Thanks in advance.

Answer 1

HINT

We have

• $x=a\sqrt[3]{b^2}\implies x^3=a^3b^2=a(ab)^2$
• $y=b\sqrt[3]{a^2}\implies y^3=b^3a^2=b(ab)^2$

and recall that $f(x)=x^3$ is strictly increasing that is

$$x_1<x_2 \iff x_1^3<x_2^3$$

Answer 2

It is $$a\sqrt[3]{b^2}<b\sqrt[3]{a^2}$$ if $$a^3b^2<b^3a^2$$ if $$a^2b^2(b-a)>0$$ and this is true.