For $a,b,c \in R$ and $a,b,c>0$. Minimize $A=a^3+b^3+c^3$

Question

For $a,b,c \in R$ and $a,b,c>0$ satisfy $a^2+b^2+c^2=27$, minimize $$A=a^3+b^3+c^3$$

you should perhaps put it in a context. Like what are the prerequisites? Do you know e.g. about Lagrange multipliers? — H. H. Rugh, Feb 25 '17 at 16:11

score 8 · Accepted Answer · answered Feb 25 '17 at 16:00

8

By Power-Mean Inequality,

$\left(\dfrac{a^3+b^3+c^3}{3}\right)^{1/3}\ge \left(\dfrac{a^2+b^2+c^2}{3}\right)^{1/2}=3$

$a^3+b^3+c^3 \ge 81$

answered Feb 25 '17 at 16:00

you can use other way ? – Word Shallow Feb 25 '17 at 16:03
@WordShallow Why do you want to use another way? This is (very) elegant, simple and efficient. – Surb Feb 25 '17 at 16:03
Do you need an alternate approach ? The other approaches would be longer – Aditya Narayan Sharma Feb 25 '17 at 16:04
2

Do you know Jensen's Inequality ? Use it for $f(x)=x^{3/2}$ – Aditya Narayan Sharma Feb 25 '17 at 16:10
yes, i do -5 more to go... – Word Shallow Feb 25 '17 at 16:15
Also with a,b,c=3 it becomes 81 so the bound is reached so 81 is the minima. – cabo Mar 01 '17 at 04:35

1 Answers1