# How to find the minimum value of this expression using AM-GM Inequality?

Consider the following problem:Let $a,b>0$ be real numbers such that $a+b\geq1$. Find the minimum value of the expression $$A=ab+{1\over ab}$$

Now using AM-GM Inequality: $$A=ab+{1\over ab}\geq 2\sqrt{ab\times \frac{1}{ab} }=2$$ However notice that the equality occurs iff $ab=1$ which contradicts the fact that $a+b\geq 1$.How should I rsolve this contradiction? Is there a way to use AM-GM Inequality to solve such kind of problems?

• How does $a b = 1$ contradict $a+b \ge 1\;$?
– dxiv
Commented Oct 21, 2016 at 5:09
• Indeed, $ab=1$ in fact implies $a+b>1$ (and hence $a+b\ge 1$) under the given assumptions, since at least one of $a$ or $b$ is at least $1$, and the other is positive. Commented Oct 21, 2016 at 5:10
• @dxiv Can you give a counter example? Commented Oct 21, 2016 at 5:12
• @tatan $a=b=1$ with $a b = 1, a+b=2\ge 1$.
– dxiv
Commented Oct 21, 2016 at 5:14
• @tatan $x \ge 2$ does not contradict $x \ge 1$. In fact, it implies it.
– dxiv
Commented Oct 21, 2016 at 5:23

For $a+b=1$ by AM-GM $$ab+\frac{1}{ab}=ab+16\cdot\frac{1}{16ab}\geq17\sqrt[17]{ab\left(\frac{1}{16ab}\right)^{16}}=17\sqrt[17]{\frac{1}{16^{16}(ab)^{15}}}\geq$$ $$\geq17\sqrt[17]{\frac{1}{16^{16}(\frac{1}{4})^{15}}}=\frac{17}{4}$$ The equality occurs for $a=b=\frac{1}{2}$
Id est, the answer is $\frac{17}{4}$.
If $a+b\geq1$ so the answer is $2$ by your AM-GM.
• Yeah this is true if $a+b\leq 1$ Commented Oct 21, 2016 at 5:41
• That's creative, but $1 \cdot 1 + \frac {1}{1 \cdot 1} = 2 \lt \frac{17}{4}$. P.S. (after the edit to the answer) Where does $a+b=1$ figure in the question?