Prove that $f(a+b)=f(a)+f(b)$

Question

I want to prove that $f(a+b)=f(a)+f(b)$ where $$f(x):= \sqrt[n]{{x}^{n-1}}$$

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I proved that $\forall x \in [0,1]$ : $f(x)>x$ and tried to demonstrate $f(a+b)=f(a)+f(b)$ starting with: let's suppose $f(a+b)>f(a)+f(b)$ ( to find a contradiction) but i didnt find any contradiction

any little hint would be appreciated

Answer 1

No, take $n=2$,

$$\sqrt 2\ne 1+1.$$