median and expectation and determining the largets find the mean and median of the pdf $f(x) = x$ for $x \in [0,\sqrt{2}]$
I got the median to be $1$ and the mean to be $\dfrac{2^{3/2}}{3}$. Is there a way of determining which is largest without a calculator - I feel it should be obvious.

To find the mean of $f$, we take $\int_0^\sqrt 2 xf(x)dx=\int_0^{\sqrt 2}x^2dx=\frac{1}{2}2^{3/2}$.
For the median, we want $\int_0^M xdx=\frac{1}{2}$, so $\frac{1}{2}M^2=\frac{1}{2}$ and $M=1$.
To see that $\frac{1}{2}2^{3/2}>1$, you can work backwards:
\begin{align}
2^{3/2}&>2\\
2^3&>2^2\\
8&>4.
\end{align}
In general, you can preserve the direction of an inequality if you are applying a strictly increasing function, such as squaring or multiplying by a positive number.
