When $abc=1$ find the minimum of $a^2+b^2+c^2$.

Question

My attempt:I put $\frac{1}{bc}$ for $a$ and then I used AM-GM inequality.

$(\frac{1}{bc})^2+b^2+c^2\ge 3$

But finally we should get the answer $a+b+c$.

Where am I wrong?

I think this will help you http://math.stackexchange.com/questions/740518/if-abc-1-then-a2b2c2-ge-abc — absolute friend, Aug 02 '16 at 12:02
It doesn't make sense to say that the minimum of $a^2+b^2+c^2$ is $a+b+c$; the minimum (or infimum) must be independent of $a,b,c$. The answer given with the problem is nonsensical. But maybe the problem asked you to show that $a^2+b^2+c^2 \ge a+b+c$, and you misinterpreted the requirement? — Alex M., Aug 02 '16 at 12:09

score 1 · Accepted Answer · answered Aug 02 '16 at 11:23

1

AM-GM says:

$$\dfrac{a^2+b^2+c^2}3\ge\sqrt[3]{a^2b^2c^2}$$

answered Aug 02 '16 at 11:23

lab bhattacharjee

I told the answer should be $a+b+c$. – Taha Akbari Aug 02 '16 at 11:49
@TahaAkbari, Why it has to be? Here the equality occurs if $a^2=b^2=c^2\implies (a,b,c)=(1,1,1)$ or $(1,-1,-1)$ – lab bhattacharjee Aug 02 '16 at 11:52
The answer is in our book. – Taha Akbari Aug 02 '16 at 11:53
@TahaAkbari if $a+b+c$ is the answer, then there is no minimum as you can arbitrarily minimize that expression – windircurse Aug 02 '16 at 11:53

score 0 · Answer 2 · answered Aug 02 '16 at 11:47

0

I think you can easily solve it using the method of lagrange multplier. And answer would be 3

answered Aug 02 '16 at 11:47

absolute friend

I thought it is 3 too. but the answer should be $a+b+c$ – Taha Akbari Aug 02 '16 at 11:51

2 Answers2