$a \geq b \geq c \geq 0$ 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.
 A: After some expanding and grouping, we have:
$$(a+b+c)^2-(a^2+3b^2+5c^2)=2(a-b)(b+c)+4c(b-c) \ge0$$
Hence the conclusion.
The equality holds if $(a-b)(b+c)=c(b-c)=0$, which surely encompasses three found equality situations.

Another solution can also be given as follows.

The set $C=\{ (a,b,c): 0 \le c \le b \le a , a+b+c \le 1 \}$ is surely convex, and in fact, it is a polyhedron formed by a finite set of points $V$ which ,in turn, serve as the vertices of the polyhedron $C$.
Besides, $f$ is a convex function on $\mathbb{R}^3$, hence the maximum of $f$ on $C$ must be achieved only in the $V$.
So what we have to do is to investigate values of $f$ on $V$. And this is straightforward, because
$$V=\{ (0,0,0) , ( 0.5,0.5,0) , (1,0,0),( 1/3, 1/3,1/3) \}$$
A: Let $ c = x$, $b = x+y$, $a = x+y+z$ where $ x, y, z \geq 0$.
Then, the question becomes:

Given non-negative $x, y,z$ such that  $ 3x+2y+z \leq 1$, show that $(x+y+z)^2 + 3 (x+y)^2 + 5 x^2 \leq 1$.

This is true because
$$(x+y+z)^2 + 3 (x+y)^2 + 5 x^2  = (3x+2y+z)^2 - (4xy+4xz+2yz) \leq 1^2 $$
We have equality iff $3x+2y+z = 1$ and  $4xy + 2xz + 2yz = 0 $.
The second condition holds when at least 2 of these non-negative terms are equal to 0, so we have the following equality cases.

*

*$ x = 0, y = 0$, we get $c = 0, b = 0 , a = 1$.

*$ x = 0, z = 0 $, we get $ c = 0, b = 1/2, a = 1/2$.

*$y = 0, z = 0 $, we get $ c = 1/3, b = 1/3, a = 1/3$.

*$x=0, y = 0, z = 0$ has no solution.


Notes

*

*I was bounding with $(3x+2y+z)^2$ because the square terms on the LHS were $ 9x^2, 4y^2, z^2$, and that expression also nicely matched with our condition.

*In hindsight, the $4xy+4xz+2yz = 2z(2x+y) + 4xy = 2(a-b)(b+c) + 4c(b-c) $ is what Paresseux had. However, IMO their factorization was hard to guess at, whereas this approach was  immediate.

A: 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.
