TITLE: Finding the maximum using inequalities
QUESTION [1 upvotes]: Given that $(a+1)(b+1)(c+1)(d+1)=81$ and $a,b,c,d>0$, find the maximum of $abcd$.
I know the answer should occur when $a=b=c=d$ but how do I get there using inequalities? This feels like AM-GM.

REPLY [2 votes]: Yeah your intuition is correct about AM-GM. But one can err badly if one naively just applies it on each factor:
$$\frac{a+1}{2}\geq \sqrt{a\cdot1}=\sqrt{a}\implies (a+1)\geq 2\sqrt{a}$$
Similarly for the other three factors and we get:
$$81 = (a+1)(b+1)(c+1)(d+1) \geq 16\sqrt{abcd}$$
$$\implies abcd \leq \bigg(\frac{81}{16}\bigg)^2\approx25.63$$
The issue is that while this bound is correct, it will never be achieved because all these equalities will never be simultaneously satisfied under the problem constraint. For each equality to be achieved we would need $a=b=c=d=1$ which violates the precondition.

But this motivates the cleverness in application of the same technique differently:
$$\frac{\frac{a}{2}+\frac{a}{2}+1}{3}\geq\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{{\frac{a}{2}\cdot\frac{a}{2}\cdot 1}}=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{{\frac{a^2}{4}}}\implies (a+1)\geq 3\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{{\frac{a^2}{4}}}$$
Similarly for the other three factors and we get:
$$81=(a+1)(b+1)(c+1)(d+1) \geq 81\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{{\frac{a^2b^2c^2d^2}{4^4}}}$$
$$a^2b^2c^2d^2 \leq 4^4 \implies abcd\leq16$$
All the equalities of AM-GM are achieved simultaneously iff $\frac{a}{2}=\frac{b}{2}=\frac{c}{2}=\frac{d}{2}=1$ equivalently iff $a=b=c=d=2$, which this time is consistent with the given constraint and thus $16$ is a tight limit, i.e. it is achieved and is the (optimal) maximum value that $abcd$ takes.
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