Compute the minimum value of $a^n + b^n + c^n$ subject to $a^2 + b^2 + c^2 = 1 $ 
Assume that $a,b,c$  are non-negative real numbers and $n$ is a natural number $n \ge 3$. What is $f(n)=$ the minimum value of $a^n + b^n + c^n$ ? 

I find ; 
$$f(3) = \frac{1}{\sqrt{3}}\qquad f(4) = \frac{1}{3}$$
then I guess 
$$f(n) = \left(\frac{1}{\sqrt{3}} \right)^n \times 3 $$ 
Is that true?

Your conjecture is true. First, $f(n)=a^n+b^n+(1-(a^2+b^2))^{\frac{n}{2}}$, calculate its derivative w.r.t. $a$, one obtains it has the minima $a=\sqrt{\frac{1-b^2}{2}}$. Inserting this into $f(n)$, we obtain $f(n)=2(\frac{1-b^2}{2})^{\frac{n}{2}}+b^n$, now calculate its derivative w.r.t. $b$, you obtain it has the minima $b=\frac{1}{\sqrt{3}}$. Now one concludes that $a=b=c=\frac{1}{\sqrt{3}}$ and $\min f(n)=3^{\frac{2-n}{2}}$.
