Find functions $f\left(x_{1}x_{2}\right)+f\left(x_{2}x_{3}\right)+\ldots+f\left(x_{n}x_{1}\right) \ge n \cdot f\left(x_{1}+x_{2}+\ldots+x_{n}\right)$

Question

Find all functions for which it exists $n$ a natural number greater or equal than $4$ for which it is true: $$f\left(x_{1}x_{2}\right)+f\left(x_{2}x_{3}\right)+\ldots+f\left(x_{n}x_{1}\right) \ge n \cdot f\left(x_{1}+x_{2}+\ldots+x_{n}\right)$$ regardless of the choice of real numbers $x_i$.

I think that this condition is simillar to a condition of convexity, therefore I should find a way to reduce the discussion to Jensen's inequality by making a renotation, for instance $y_1 = x_1x_2$, $y_2 = x_2x_3$ etc. However I didn't manage to find a way to reach a result.

Answer 1

The condition looks like a convexity condition, but as we'll see, it is much stronger: the only functions satisfying the condition are constant.

If we set $x_1 = d$, $x_3=s-d$, and $x_i = 0$ for $i \notin \{1, \ldots, n\}$, then the left hand equals $nf(0)$ and the right hand side equals $nf(s)$, so the inequality reduces to $f(0) \geq f(s)$, where $s$ is arbitrary.

Now if we take $x_2 = -x_1$ and $x_i = 0$ or $i \notin \{1, 2\}$, then the right hand side becomes $n f(0)$ and the left hand side becomes $f(-x_1^2) + (n-1)f(0)$. The inequality then reduces to $f(-x_1^2) \geq f(0)$. Since $-x_1^2$ can take any value $\leq 0$, we find that for all $s \leq 0$, $f(s) \geq f(0) \geq f(s)$, i.e. $f(s) = f(0)$.

Now set $x_1=x_2$ and $x_3=x_4=-x_1$, with $x_i=0$ for $i \notin \{1,2,3,4\}$. For $n \geq 5$, the inequality becomes $$f(x_1^2)+f(-x_1^2)+f(x_1^2)+(n-3)f(0) \geq nf(0)$$ using that $f(-x_1^2)=f(0)$ from above, this reduces to $2f(x_1^2) \geq 2f(0)$, i.e. $f(x_1^2) \geq f(0)$. If instead $n=4$, the inequality becomes $$f(x_1^2)+f(-x_1^2)+f(x_1^2)+f(-x_1^2) \geq 4f(0)$$ Using that $f(-x_1^2)=f(0)$ again, we see that this also reduces to $2f(x_1^2) \geq 2f(0)$, i.e. $f(x_1^2) \geq f(0)$. In either case $f(x_1^2) \geq f(0)$. Since $x_1^2$ can take any nonnegative value, we see that for any nonnegative $s$, $f(s) \geq f(0)$. Coupled with the above, this gives $f(s) \geq f(0) \geq f(s)$, so $f(s)=f(0)$.

We've now proven that if $f$ satisfies the given condition, then $f(s) = f(0)$ for all $s \in \mathbb R$. This means that only the constant functions can satisfy the condition.

If $f(x) \equiv C$, then the inequality becomes $nC \geq nC$, so constant functions satisfy the condition. Thus, the set of functions satisfying the given conditions is precisely the set of constant functions.