falantrng wrote: Let $\mathbb R$ be the set of real numbers. Determine all functions $f:\mathbb R\to\mathbb R$ that satisfy the equation $$\sum_{i=1}^{2015} f(x_i + x_{i+1}) + f\left( \sum_{i=1}^{2016} x_i \right) \le \sum_{i=1}^{2016} f(2x_i)$$for all real numbers $x_1, x_2, ... , x_{2016}.$ If $f(x)$ is a solution, so is $f(x)+a$, whatever is $a\in\mathbb R$. So WLOG $f(0)=0$ Using $(x,-x,x,-x,...,x,-x)$ in FE, we get $f(x)+f(-x)\ge 0$ Using $(x,x,0,0,...,0,0)$ in FE, we get $ff(x)\le 0$ And so $f\equiv 0$ and, back to WLOG : $\boxed{f(x)=c\quad\forall x}$, which indeed fits, whatever is $c\in\mathbb R$

1. Choose \( x_1 = x \), \( x_2 = x \), and \( x_i = 0 \) for \( i \geq 3 \). This gives us \[ 2f(2x) + f(x) + 2013f(0) \leq 2f(2x) + 2014f(0), \]which implies \[ \boxed{f(x) \leq f(0) \text{ for all } x \in \mathbb{R}} \tag{*} \] 2. Secondly, choose \[ x_i = \begin{cases} \frac{x}{2} & \text{if } i \text{ is odd}, \\ -\frac{x}{2} & \text{if } i \text{ is even}. \end{cases} \]We get \[ 2015 f(0) + f(0) \leq 1008f(x) + 1008 f(-x), \]implying \[ \boxed{2f(0) \leq f(x) + f(-x) \text{ for all } x \in \mathbb{R}} \tag{**} \] Combining \((*)\) and \((**)\), we have \[ 2f(0) \leq f(x) + f(-x) \leq 2f(0). \]Hence, \[ 2f(0) = f(x) + f(-x), \]call this \((***)\). Again by combining \((*)\) and \((***)\), we have \[ 2f(x) \leq 2f(0) = f(x) + f(-x). \]This implies \[ f(x) \leq f(-x) \text{ for all } x \in \mathbb{R}. \]By replacing \( x \) with \(-x\), we obtain \( f(x) = f(-x) \) for all \( x \in \mathbb{R}. \) Plugging that into \((***)\), we get \[ f(x) = f(0) = k, \text{ a constant, for all } x \in \mathbb{R}. \]Plugging back into the original equation, we see that this is indeed the only solution.