The maximum between two real numbers at denominator

Question

In writing a proof I have encountered a problem that I hope to solve here.
If I have in general a term like $$\frac{1}{\max\{a,b\}}$$ where $a,b\in\mathbb{R}\setminus\{0\}$ then it is right to say the following? $$\frac{1}{\max\{a,b\}}\leq \min\left \{\frac{1}{a}, \frac{1}{b}\right \}$$

Answer 1

The inequality is false. Take, for example, $a=1$ and $b=-1$.

Then $$\frac{1}{\max\{a, b\}} = \frac{1}{\max\{1, -1\}} = \frac{1}{1}=1$$ and $$\min\left\{\frac{1}{a},\frac{1}{b}\right\} = \min\left\{\frac{1}{1},\frac{1}{-1}\right\} = \min \{1,-1\} = -1$$

Answer 2

There will be three cases possible:

1. a and b both are positive.
   Without any loss of generality, we can assume a to be smaller than b. Then, in that case, max{a,b} = $\frac{1}{b}$ and $min\{\frac{1}{a},\frac{1}{b}\}$ = 1/b. Therefore, $\frac{1}{max\{a,b\}}$ = $min\{\frac{1}{a},\frac{1}{b}\}$. Equality holds.
2. among a and b, one is positive and other is negative.
   Again, without any loss of generality, we can assume a is negative. Then, in that case, max{a,b} = b and $min\{\frac{1}{a},\frac{1}{b}\}$ = $\frac{1}{a}$. Therefore, $\frac{1}{max\{a,b\}}$ > $min\{\frac{1}{a},\frac{1}{b}\}$, as a is negative. Hence the give equality fails.
3. a and b both are negative.
   Without any loss of generality, we can assume a to be smaller than b. Then, in that case, max{a,b} = b and $min\{\frac{1}{a},\frac{1}{b}\}$ = $\frac{1}{b}$. Therefore, $\frac{1}{max\{a,b\}}$ = $min\{\frac{1}{a},\frac{1}{b}\}$, as both are negative. Equality holds.

Hence, the given equality holds only when both the numbers are either positive or negative.