Minimizing the sum of rational functions over simplex

Question

Let

$$f(x) := \frac{2x+1}{x^2+3}$$

I would like to solve the following optimization problem

$$ \begin{array}{ll} \underset {x_1, x_2, x_3} {\text{minimize}} & f(x_1) + f(x_2) + f(x_3) \\ \text{subject to} & x_1 + x_2 + x_3 = 3 \\ & x_1, x_2, x_3 \in [0,3] \end{array} $$

I am not sure how to find that since $f(x)$ is concave and convex over the given interval. Any help will be greatly appreciated.

Answer 1

We have, for all $x \in [0, 3]$, $$\frac{2x + 1}{x^2 + 3} - \frac{x + 4}{12} = \frac{x(x + 7)(3 - x)}{12(x^2 + 3)} \ge 0.$$

Thus, we have, for all $x_1, x_2, x_3 \in [0, 3]$ with $x_1 + x_2 + x_3 = 3$, $$f(x_1) + f(x_2) + f(x_3) \ge \frac{x_1 + 4}{12} + \frac{x_2 + 4}{12} + \frac{x_3 + 4}{12} = \frac54$$ with equality if $x_1 = 0, x_2 = 0, x_3 = 3$ and permutations.

Thus, the minimum if $\frac54$.

Answer 2

Uhhh. I didn't realize that River Li was just using tangent line trick...I had nothing to do so I present a proof that if probably wrong but I can't find why....
Let $f(x) = \frac{2x+1}{x^2+3}$. Therefore, $f'(x) = \frac{-2x^2-2x+6}{(x^2+3)^2}$. Therefore $f''(x) = \frac{4x^3+6x^2-36x-6}{(x^2+3)^3}$
We are now ready to locate the inflection points of $f(x)$ $$f''(x) = 0 \leftrightarrow 4x^3+6x^2-36x-6 = 0$$ Since \begin{array}{|c|c|} \hline x& f''(x)\\ \hline -4&-22 \\ \hline -3&48 \\ \hline 0&-6 \\ \hline 3&48 \\ \hline \end{array} And clearly $-4,-3,0,3$ are not zeros. From intermediate value theorem follows that there is only one root of the cubic in the interval $[0,3]$ and thus one inflection point ot $f(x)$.

We will now use something known as n - 1 EV (Theorem 2.8) $\leftarrow$ Please someone check if I am using this correctly

We now just need to optimise: $F(a) =f(a) + f(a) + f(3-2a)$ subject to $a\in[0,1.5]$

$$F(a) = 2*\frac{2a+1}{a^2+3}+\frac{2(3-2a)+1}{(3-2a)^2+3}\Longrightarrow F'(a) = \frac{4a^2-14a+9}{4(a^2-3a+3)^2}-4*\frac{a^2+a-3}{(a^2+3)^2}$$ We will now find the critical points in $F'(a)$ subject to $a\in[0,1.5]$

Since $a\in \mathbb{R}$ the function is always defined and we will only need to check $F'(a) = 0$ $$F'(a) = 0 \leftrightarrow \frac{4a^2-14a+9}{4(a^2-3a+3)^2}=4\frac{a^2+a-3}{(a^2+3)^2}$$ $$12a^6-66a^5+63a^4+324a^3-954a^2+1134a-513=0$$ From some testing we get the only real root in the interval is $a = 1$ (I did Sturm's method with wolfram if anyone knows a method to do this by hand please please reply ;D )

Now onto our last step:

\begin{array}{|c|c|} \hline a& F'(a)\\ \hline 0&\frac{5}{4} \\ \hline 1&\frac{9}{4} \\ \hline 1.5&1.85714 \\ \hline \end{array}

And finally we conclude that the minimum is $\frac{5}{4}$ with the equality cases being permutations of (0,0,3)

And also the maximum reached at (1,1,1) (Can also be found using concave Jensen)