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?
 A: 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 $\min f(n)=3^{\frac{2-n}{2}}$.
A: Set $(x_1, x_2, x_3) = (a, b, c)$. Applying Holder's inequality to $(x_1^2, x_2^2, x_3^2)$ and $ (1, 1, 1)$
with 
$$
 p = \frac n2 \, , \quad q = \frac{n}{n-2}
$$
gives
$$
 1 = \sum_{k=1}^3 x_k^2 \cdot 1 \le \left( \sum_{k=1}^3 x_k^n \right)^{\frac 2n} \cdot  \left( \sum_{k=1}^3 1 \right)^{\frac {n-2}{n}} \\
 =  \left( \sum_{k=1}^3 x_k^n \right)^{\frac 2n} \cdot  3 ^{\frac {n-2}{n}}
$$
which implies
$$
 x_1^n + x_2^n + x_3^n\ge \left(\frac 13 \right)^{\frac {n-2}{2}} = 
 \frac{3}{(\sqrt 3)^n}
$$
and equality holds for $x_1  = x_2 = x_3 = \frac{1}{\sqrt 3}$,
so this is the minimum value, as you conjectured.
A: The power mean inequality gives $$\sqrt[n]{\frac{a^n+b^n+c^n}{3}} \geq \sqrt[2]{\frac{a^2+b^2+c^2}{3}}=\sqrt{\frac13}$$
with equality if $a=b=c$. Hence the minimal value is $3\left(\frac13\right)^{n/2}$, which is attained if $a=b=c=\left(\frac13\right)^{1/2}$. 
A: The answer is true. Once one has proved that the minimum is achieved for $a=b=c$ (which can be done for example with standard Lagrange multiplier metohd), then the value is easily obtained:
$$
\min (f)=3 a^n,\;1=3a^2\Rightarrow \min (f)=3^{1-n/2}
$$
