Simple Inequality, maximisation.

Question

Maximise $(a+1)(b+1)(c+1)$ when $a+b+c=1$.

I know the answer is just set $a=b=c=\frac{1}{3}$, and you could prove that by comparing two terms at a time: fix one term, and the two other terms are maximised when equal because of squares, so you need to set all 3 terms equal.

But is there any proper solution to this using real inequality techniques?

The most immediate method that comes to mind is that of Lagrange multipliers. — PrincessEev, Jun 09 '23 at 09:09
The maximum does not exist. For non-negative variables use AM-GM — Michael Rozenberg, Jun 09 '23 at 09:19
&PrincessEev This is one of very many methods, which we can use here. LM here is one of bad ways for the solution — Michael Rozenberg, Jun 09 '23 at 10:04

score 2 · Answer 1 · answered Jun 09 '23 at 15:09

The maximum value does not exist.

Try $a=b\rightarrow-\infty$.

For non-negative variables by AM-GM we obtain: $$\prod_{cyc}(a+1)\leq\left(\frac{\sum\limits_{cyc}(a+1)}{3}\right)^3=\frac{64}{27}.$$ The equality occurs for $a=b=c=\frac{1}{3},$ which says that we got a maximal value.

Simple Inequality, maximisation.

1 Answers1