Maximize $\frac{a^2b}{(1+a+ab)^3}$ subject to $a,b\ge1$

Question

Note: The two problems are basically equivalent, but I wonder if one derives it back through the same path or maybe solves it in an easier way.

Do you know of the arithmetic-geometric mean inequality? – πr8 May 20 '17 at 11:51 — πr8, May 20 '17 at 11:51

score 0 · Accepted Answer · answered May 20 '17 at 11:48

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HINT: Show that $$\frac{a^2b}{(1+a+ab)^3}\le \frac{1}{27}$$ for $$a,b\geq 1$$

answered May 20 '17 at 11:48

Yes, by AM-GM: $(1+a+ab)^3\ge (3\sqrt[3]{1\cdot a\cdot ab})^3$ – farruhota May 20 '17 at 11:59
why the $-1$? I don't know, what is wrong? – Dr. Sonnhard Graubner May 21 '17 at 07:34

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