Minima and maxima of $ab+\frac{1}{ab}$

Question

Given:

$ab+\frac{1}{ab}$

$a,b>0$

$a+b\leq1$

Using the AM-GM & doing some manipulations I got, $0<ab\le \frac{1}{4} $.

then $\frac{1}{ab}>4$. Now my idea is to simply add both inequalities $ab,\frac{1}{ab}$.

But those sign of the inequalities aren't same!! So what can I do!!

Answer 1

By AM-GM we obtain $ab\leq\frac{1}{4}$ and since $$\left(x+\frac{1}{x}\right)'=1-\frac{1}{x^2}<0$$ for $0<x<\frac{1}{4}$ we obtain: $$ab+\frac{1}{ab}\geq\frac{1}{4}+\frac{1}{\frac{1}{4}}=\frac{17}{4}.$$ The equality occurs for $a=b=\frac{1}{2},$ which says that we got a minimal value.

The maximal value does not exist. Try $b\rightarrow0^+$.

Answer 2

You observe that in $x+1/x$ for any positive x, minima is at x= 1, that is as we move away from 1, the value increases. Here, you have ab in place of x. the expression should be minimum at ab=1, but it is not allowed. The closest value allowed is ab=1/4. So minima is at =1/4. As for the upper bound, you can clearly see that it doesn't exist

Note: this kind of analysis is simple only because the derivative of function is zero only at one point for X>0