Find $f(10)=?$ when the following condition is given. $$ Let \ f:R \to R \in \vert f(x)-f(y) \vert \le (x-y)^3 \ \forall \ x,y \in R \\and \ f(2)=5;\ then\ f(10)=? $$
This question is from an old assignment on the topic Limits , Continuity and Differentiability
Though  i didn't get  the answer , but i tried in the following way.... 
    $$ \vert f(x)-f(y) \vert \le (x-y)^3 \ \\ \implies 0\le \vert f(x)-f(y) \vert \le (x-y)^3 \\ \implies x-y\ge0 \implies x\ge y\ \forall (x,y) \in domain \\ put\ x=10 \ and \ y=2 \implies|f(10)-5| \le 8^3 \\ \implies \ f(10) \in (-8^3+5,8^3+5) $$ But i couldn't get any further. Please help.( I think we need to use squeeze theorem) 
$$ **EDIT** (previous\ inequality \ is\\ wrong\ \mathbf| answer \ as \ helped\ by\ \mathbf{Przemysław Scherwentke}\ and\ \mathbf{ mengdie1982} ) \\\vert f(x)-f(y) \vert \le |x-y|^3 \\-|x-y|^2 \leq \frac{f(x)-f(y)}{x-y}\leq |x-y|^2, ~~~\forall x \neq y.\\ 0\leq\lim_{y \to x}\frac{f(x)-f(y)}{x-y} \leq 0,\implies f'(x) = 0,~~~\forall x \in \mathbb{R}.\\ hence\ f(10)=5 $$

Maybe, the inequality condition of the problem should be $$|f(x)-f(y)|\leq |x-y|^3,~~\forall x,y \in \mathbb{R}.$$
Thus, $$-|x-y|^2 \leq \frac{f(x)-f(y)}{x-y}\leq |x-y|^2, ~~~\forall x \neq y.$$
Now, fix $x$ and take all the limits under the process $y \to x.$ We may obtain $$0\leq\lim_{y \to x}\frac{f(x)-f(y)}{x-y} \leq 0,$$ which implies $$f'(x) = 0,~~~\forall x \in \mathbb{R}.$$Hence, $f(x)$ is a constant function. As a result, $$f(10)=f(2)=5.$$
