Rooting for oneself?

First, compare $\sqrt[2]{2}$ and $\sqrt[3]{3}$ by raising each to the sixth power: $(\sqrt[2]{2})^6 = 2^3 = 8;\ \ (\sqrt[3]{3})^6 = 3^2 = 9;\ \ \text{since}\ 8 < 9,\ \sqrt[2]{2} < \sqrt[3]{3}.$ Next, compare $\sqrt[3]{3}$ and $\sqrt[4]{4}$ by raising each to the twelfth power: $(\sqrt[3]{3})^{12} = 3^4 = 81;\ \ (\sqrt[4]{4})^{12} = 4^3 = 64;\ \ \text{since}\ 81 > 64,\ \sqrt[3]{3} > \sqrt[4]{4}.$ (Or save yourself some work by noticing $\sqrt[2]{2} = \sqrt[4]{4}$ .)

Finally, compare $\sqrt[3]{3}$ and $\sqrt[5]{5}$ by raising each to the fifteenth power: $(\sqrt[3]{3})^{15} = 3^5 = 243;\ \ (\sqrt[5]{5})^{15} = 5^3 = 125;\ \ \text{since}\ 243 > 125,\ \sqrt[3]{3} > \sqrt[5]{5}.$

Thus $\boxed{\sqrt[3]{3}}$ is the biggest of the bunch.

Let $f(x)=x^\frac{1}{x}$ We need to compare $f(2), f(3) , f(4) , f(5)$

$\ln(f(x))=\frac{\ln(x)}{x}$ Differentiating, $\frac{f'(x)}{f(x)}=\frac{x/x-\ln(x)}{x^2}$

For maximum, $f'(x)=0$

$1-\ln(x)=0$

So maximum value of $f(x)$ is at $x=e$

So, we can for sure say that f(3)>f(4)>f(5) so we need to only compare f(2) and f(3).

If you do not want to compare, just see that 3 is much closer to $e$ than 2. This does not work always so better to check but the answer is 3

First, convert each radical to its fractional exponent form:

$\sqrt[2]{2} = 2^{\frac{1}{2}}$
$\sqrt[3]{3} = 3^{\frac{1}{3}}$
$\sqrt[4]{4} = 4^{\frac{1}{4}}$
$\sqrt[5]{5} = 5^{\frac{1}{5}}$

Now compare $2^{\frac{1}{2}}$ and $4^{\frac{1}{4}}$ . With some simplification you can notice that they are equal.

$2^{\frac{1}{2}}=4^{\frac{1}{4}}$

Now compare $3^{\frac{1}{3}}$ and $5^{\frac{1}{5}}$ . Notice that $3^3$ and $5^2$ are almost equal. We can use this fact to compare the two values.

Let $x=3^{\frac{1}{3}}$ and $y=5^{\frac{1}{5}}$ . This means $x^9=3^3$ and $y^{10}=5^2$ . As $x^9$ and $y^{10}$ are now almost equal, it is clear that $x>y$ as $x$ has a smaller exponent. Therefore,

$3^{\frac{1}{3}}>5^{\frac{1}{5}}$ .

Now you could compare $2^{\frac{1}{2}}$ and $3^{\frac{1}{3}}$ , however each answer choice only contains one option so it is clear that the answer must be $\boxed{3^{\frac{1}{3}}}$ as $2^{\frac{1}{2}}=4^{\frac{1}{4}}$ .