If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$? If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?
I tried with Tchebyshev inequality on sets $\{a, b, c\}$ and $\{a^2, b^2 , c^2\}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.
I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.

You can do it using the mean of the $m$ power for $m=\frac{3}{2}$ (or convexity of $x^\frac{3}{2}$ for that matter) :
We have that $\frac{x^{\frac{3}{2}}+y^{\frac{3}{2}}+z^\frac{3}{2}}{3} \geq (\frac{x+y+z}{3})^\frac{3}{2}$ for any $x,y,z > 0$ 
Substituting $x=a^2, y=b^2, z=c^2$ we obtain $\frac{a^3+b^3+c^3}{3} \geq (\frac{a^2+b^2+c^2}{3})^\frac{3}{2} = 9^\frac{3}{2}=27$ and therefore the sought minimum is $3\cdot 27= 81,$, attainable if(and only if) all three numbers are equal to $3$.
