In my lecture notes, it is claimed with no justification that the solution of the optimization problem
$$\begin{align}
\begin{split}
\min_{x_1,\ldots,x_n\in\mathbb R}\ \ &\sum_{i=1}^n x_i^5 \\
\text{s.t.}\ \ &\sum_{i=1}^n x_i =1\\
&x_1,\ldots,x_n\ge 0
\end{split}\end{align} $$
is given by $x_1^*=\ldots=x_n^*=\frac{1}{n}$.
It is not obvious to me, so I tried to prove it by computing the Lagrangian and writing down the KKT conditions, which are
$$\begin{aligned}
L(x, \mu, \lambda) &= \sum_{i=1}^n x_i^5 + \mu(\sum_{i=1}^n x_i – 1) +\sum_{i=1}^n \lambda_i (-x_i)\\
5x_i^4 + \mu – \lambda_i &= 0 & (\nabla L = 0) \\
\lambda_i (-x_i) &= 0 & \text{(Complementarity)} \\
\sum_{i=1}^n x_i = 1,~ x_i &\geq 0 & \text{(Primal feasibility)} \\
\lambda &\geq 0 &\text{(Dual feasibility)}
\end{aligned} $$
My Lagrangian optimization is quite rusty, but I think that it's possible to argue as follows : if there is one $x_j=0$, first order condition implies that $\mu = \lambda_j$. Furthermore the feasibility condition also implies that there must be at least one $x_k>0$. We can rewrite the Lagrangian as
$$ L(x, \mu, \lambda) = \sum_{i=1}^n x_i^5 + \sum_{i=1\\i\ne j}^n (\lambda_j-\lambda_i)x_i – \lambda_j$$
Now I would like to say that since by complementarity, $\lambda_k = 0$, we can find values of $\lambda_j$ which make the Lagrangian blow up, but I'm a bit stuck with formalizing it… How to finish ? Also, is there an easier way to prove it ?
I noticed that the problem is equivalent to finding the probability distribution over $\{1,\ldots,n\} $ with minimal fourth moment, but I don't see how that's helpful.
Best Answer
You can prove it, for example, by Jensen's inequality with the convexe function $f(x) = x^5$ $$\frac{\sum_{i=1}^5 x_i^5}{n} \ge \left( \frac{\sum_{i=1}^n x_i}{n}\right)^5$$ The equality occurs if and only if $x_1=..=x_n=\frac{1}{n}$