Why AM-GM inequality showing different results?

Question

I was given to find the minimum of $$1+a_1+a_2+a_3+...a_n$$. It was given that $$a_1\times{a_2}\times{a_3}...a_n =c$$ My approach


Using AM-GM inequality
$1+a_1+a_2+...a_n\ge (n+1)c^{\frac{1}{n+1}}$


But the story doesn't ends here. There could be one more approach.

$a_1+a_2+a_3+...a_n\ge nc^{\frac{1}{n}} $ So, $1+a_1+a_2+...a_n\ge 1+nc^{\frac{1} {n}} $


Both the approaches are giving different results. Is there something that I am missing?

Answer 1

Both results are correct, but one of them tells you more than the other. For instance, if $x=20$, then both $x\ge10$ and $x\ge 5$ are correct, but $x\ge 10$ tells you more.

In your case, the second approach tells you more, because it uses explicitly the knowledge that the first term is equal to $1$. In fact, with a bit of work, you could even use this to prove that $1+nc^\frac{1}{n}\ge(n+1)c^\frac{1}{n+1}$.

Answer 2

Both results are correct, but they have different conditions for when the equality holds.

For the first inequality, the equality holds when $1 = a_1 = \cdots = a_n$. But for the second inequality, the equality holds when $a_1=\cdots = a_n$.

As TonyK mentions above, the second inequality tells you more. The equality condition would constrain all the $a_i$s and causes $c = 1 (=a_1\dots a_n)$. But the second condition for the equality does not constrain any values for $c$.