How to determine without calculator which is bigger, $\left(\frac{1}{2}\right)^{\frac{1}{3}}$ or $\left(\frac{1}{3}\right)^{\frac{1}{2}}$ How can you determine which one of these numbers is bigger (without calculating):
$\left(\frac{1}{2}\right)^{\frac{1}{3}}$ , $\left(\frac{1}{3}\right)^{\frac{1}{2}}$
 A: Raise them both to the power of $6$.
Since they are both positive, their order will be preserved and you will get:
$$\left({\dfrac{1}{2}}\right)^2=\frac{1}{4} > \frac{1}{27}=\left({\dfrac{1}{3}}\right)^3$$
A: $((\frac{1}{2})^{\frac{1}{3}})^6=(\frac{1}{2})^2=\frac{1}{4}$
$((\frac{1}{3})^{\frac{1}{2}})^6=(\frac{1}{3})^3=\frac{1}{27}$
So as it is obvious from the above relations, $((\frac{1}{2})^{\frac{1}{3}})^6>((\frac{1}{3})^{\frac{1}{2}})^6$, so we can say $(\frac{1}{2})^{\frac{1}{3}}>(\frac{1}{3})^{\frac{1}{2}}$
A: If you don't want to depend on the "trick" of raising to the sixth power, you can compare the logs:  $\frac 13 \log \frac 12=\frac {- \log 2}3$ and $\frac 12 \log \frac 13=\frac {-\log 3}2$ Now $\frac 12 \gt \frac 13$ and $\log 3 \gt \log 2$, so $\frac {\log 3}2 \gt \frac {\log 2}3, \frac {-\log 3}2 \lt \frac {-\log 2}3,\left(\frac{1}{3}\right)^{\frac{1}{2}} \lt \left(\frac{1}{2}\right)^{\frac{1}{3}}$ 
A: $$\left(\frac{1}{2}\right)^{\frac{1}{3}}=\frac{\sqrt[3]1}{\sqrt[3]2}=\frac1{\sqrt[3]2}$$
$$\left(\frac{1}{3}\right)^{\frac{1}{2}}=\frac{\sqrt1}{\sqrt3}=\frac1{\sqrt3}$$
Now it is obvious that
$$\sqrt[3]2<\sqrt3$$
Thus
$$\frac1{\sqrt[3]2}>\frac1{\sqrt3}$$
A: No need to do any calculations at all: since we are talking about numbers between $0$ and $1$, a cube root is larger than a square root:
$$\Bigl(\frac12\Bigr)^{1/3}>\Bigl(\frac12\Bigr)^{1/2}>\Bigl(\frac13\Bigr)^{1/2}\ .$$
A: Since $3^3=27 > 4=2^2$, we have
$$
\left(\frac{1}{2}\right)^2
>
\left(\frac{1}{3}\right)^3
$$
Take square roots and get
$$
\left(\frac{1}{2}\right)
>
\left(\frac{1}{3}\right)^{3/2}
$$
Take cube roots and get
$$
\left(\frac{1}{2}\right)^{1/3}
>
\left(\frac{1}{3}\right)^{1/2}
$$
A: For $x ≥ 0$ and $y > 0$, $x^y$ grows with $x$ for fixed $y$; it grows with $y$ for fixed $x > 1$, and decreases with $y$ for fixed $x < 1$. This becomes much more obvious if you take the logarithm. Special case as David said: For $0 < x < 1$, a cube root is greater than a square root. 
So increasing $x$ from $1/3$ to $1/2$ increases the result, and decreasing the exponent from $1/2$ to $1/3$ also increases the result since $x < 1$. No need to calculate anything. 
Checking whether $\left(\frac{1}{3}\right)^{1/3} < \left(\frac{1}{2}\right)^{1/2}$ would have been more difficult. Raising to the sixth power would give $1/9 < 1/8$ which is a lot closer than $1/4 > 1/{27}$. 
A: When is $x^y > y^x$ ?
When $x^{1/x} > y^{1/y}$.
Let's look at the function $x^{1/x}$.
Differentiating, we find it has a maximum at $x=e$.
Since $1/2$ and $1/3$ are both less than $e$, the one that's nearer wins.
So $(1/2)^2 > (1/3)^3$,
so $(1/2)^{1/3} > (1/3)^{1/2}$.
But more to the point, this shows that $e^\pi > \pi^e$,
which might be a lot harder without a calculator.
A: What's wrong with just logging? Logarithm is a monotone increasing function, so the inequality sign stays the same. 
First log, the multiply both sides by 2 and 3. LHS becomes $\log (\frac{1}{2})^2$, right $\log(\frac{1}{3})^3$. Now exponentiate. LHS is $\frac{1}{2} \cdot \frac{1}{2} \cdot 1$, RHS is $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3}$. Each term on the LHSis greater than each term on the RHS, hence the inequality follows. 
