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.
 A: Okay, the question requires knowledge of the AM-GM inequality, which you do have.
You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
The answer is $81$.
Hope this helps.
A: 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 $3$.
A: From the power mean inequality we know that 
$$\sqrt[q]{\sum_{i=1}^nw_ix_i^q}  \geq \sqrt[p]{\sum_{i=1}^nw_ix_i^p} \quad \forall \ \ p <q$$
where $ \sum_{i=1}^n w_i=1 $
In your case $p=2, q=3$ and $n=3$. Hence,
$$\left(\dfrac{a^3+b^3+c^3}{3}\right)^{1/3}\ge \left(\dfrac{a^2+b^2+c^2}{3}\right)^{1/2}=3$$
A: For $a=b=c=3$ we obtain a value $81$.
We'll prove that it's a minimal value.
Indeed, we need to prove that $$a^3+b^3+c^3\geq81,$$ which is true because
$$a^3+b^3+c^3-81=\sum_{cyc}(a^3-27)=$$
$$=\sum_{cyc}\left(a^3-27-\frac{9}{2}(a^2-9)\right)=\frac{1}{2}\sum_{cyc}(a-3)^2(2a+3)\geq0.$$
Also, Holder helps:
$$a^3+b^3+c^3=\sqrt{\frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}\geq\sqrt{\frac{(a^2+b^2+c^2)^3}{3}}=81.$$
