$a \geq b \geq c \geq 0$ and $a + b + c \leq 1$. Prove that $a^2 + 3b^2 + 5c^2 \leq 1$. Positive numbers $a,b,c$ satisfy $a \geq b \geq c$ and $a + b + c \leq 1$. Let $f(a, b, c) = a^2 + 3b^2 + 5c^2$. Prove that $f(a, b, c) \leq 1$.
One observation is that the bound is met: $f(1, 0, 0) = f\left(\frac{1}{2}, \frac{1}{2}, 0\right) = f\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right) = 1$.
Another observation is that clearly $a + b + c = 1$ at a maximum of $f$ since increasing any of $a,b,c$ increases $f$. So I'm just going to assume that $a + b + c = 1$ from now on.
My progress so far is that using Lagrange multipliers, you can see that $f$ is minimised subject to the constraint $a + b + c = 1$ at the point $\left(\frac{15}{23}, \frac{5}{23}, \frac{3}{23}\right)$. And since there are no other minimums it must be increasing as you choose $(a, b, c)$ away from this point. Setting $a = 1 - b - c$, we just need to optimise in $b, c$. The optimisation is bounded by the triangle formed from the lines $$c = 0, b = \frac{1}{2} - \frac{1}{2}c, \textrm{ and } c = b.$$
As $f$ is increasing as you move away from the minimum, the optimisation occurs at one of the corners of this triangle, and these give the three solutions that I gave at the start.
I think my

it must be increasing as you choose $(a, b, c)$ away from this point.

claim is a bit vague and I'm not sure how to formulate it properly. Hopefully there is a clearer method of proving this.
I found the puzzle here in case it's of interest.

Suppose we are not at one of the three vertices of the triangle, and note that the sides of the triangle correspond to $c=0$, $b=c$ and $a=b$. This means we must have at least one of the following:

*

*$a>b$ and $c>0$

*$a>b$ and $b>c$

*$b>c$ and $c>0$
In all of these cases we necessarily have $a<1$.
In the first case, choose $\delta>0$ sufficiently small. It's easy to check that $$f(a+2\delta,b-\delta,c-\delta)+f(a-2\delta,b+\delta,c+\delta)>2f(a,b,c),$$
so one of these two modifications increases the function (while remaining feasible) and so $f(a,b,c)$ is not maximal.
In the other two cases you can do the same by considering $(a\pm\delta,b\mp\delta,c)$ and $(a\pm\delta,b\pm\delta,c\mp2\delta)$ respectively.
