This is a neater version of a question I posted here, on which I'm also stuck.

**The problem:** Say I have a probability density function $f(x)$, defined for positive $x$, and let's note its $n$th non-centred moment $x_{n}$. Now, if the mean $x_{1}$ is fixed, what $f(x)$ minimises the second moment $x_{2}$? My intuition says it should be an exponential distribution (or similar shape).

**My attempt:** I want to minimise the functional

$J(y)=\int x^{2}f(x)dx$

with the constraints of mean and unit integral:

$\int xf(x)dx=x_{1}$

$\int f(x)dx=1$

So I need to minimise the modified functional:

$J^{\ast}(y)=\int x^{2}f(x)dx+\alpha \int xf(x)dx + \beta \int f(x)dx$

$J^{\ast}(y)=\int (x^{2}f(x) +\alpha xf(x) + \beta f(x))dx$

$J^{\ast}(y)=\int (x^{2} +\alpha x + \beta )f(x)dx$

Applying Euler-Lagrange gives $x^{2}+\alpha x +\beta =0$, where $f(x)$ doesn't appear!

So I don't know how to do, especially for finding the values of $\alpha$ and $\beta$. I also tried adding a constraint of positivity on $f(x)$, by substituting it with $u(x)^{2}$ (always positive by construction), but it leads to the same Euler-Lagrange equation.

Could anyone help me on that? Many thanks!