<s>[INST] 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.
 [/INST] 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:
$$a^3+b^3+c^3=\sqrt{\frac{(a^3+b^3+c^3)^2(1+1+1)}{3}}\geq\sqrt{\frac{(a^2+b^2+c^2)^3}{3}}=81.$$
</s>