Vtyjkyuk

$$ Let f:R to R in vert f(x)-f(y) vert le (x-y)^3 forall x,y in R \and f(2)=5; then f(10)=? $$
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....
$$ vert f(x)-f(y) vert le (x-y)^3 \ implies 0le vert f(x)-f(y) vert le (x-y)^3 \ implies x-yge0 implies xge y forall (x,y) in domain \ put x=10 and y=2 implies|f(10)-5| le 8^3 \ implies f(10) in (-8^3+5,8^3+5) $$ But i couldn't get any further. Please help.( I think we need to use squeeze theorem)
$$ **EDIT** (previous inequality is\ wrong mathbf| answer as helped by mathbfPrzemysław Scherwentke and mathbf mengdie1982 ) \vert f(x)-f(y) vert le |x-y|^3 \-|x-y|^2 leq fracf(x)-f(y)x-yleq |x-y|^2, ~~~forall x neq y.\ 0leqlim_y to xfracf(x)-f(y)x-y leq 0,implies f'(x) = 0,~~~forall x in mathbbR.\ hence f(10)=5 $$

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

HINT: Show at the beginning that $f'(x)equiv0$.