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?
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.