Prove or disprove: If $a+b \leq \frac{1}{2}$, then $\frac{(1-a)(1-b)}{ab} \geq 1$ for positive $a,b$

Question

Let $a,b$ be two positive numbers. Prove or disprove the statement:

If $a+b \leq \frac{1}{2}$, then $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$.

True. Assume $a+b \leq \frac{1}{2}$. Then

$$\dfrac{1-a}{a} \dfrac{1-b}{b}=\dfrac{1-b-a+ab}{ab}=\dfrac{1}{ab}-\dfrac{a+b}{ab}+1=-\dfrac{a+b}{ab}+\dfrac{1}{ab}+1\geq \dfrac{-1}{ab}+\dfrac{1}{ab}+1=1. $$

Can you check my answer?

Answer 1

You have succeeded in proving that $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$ if $a,b>0$ and $a+b \leq 1.$

Therefore, since $\frac12<1$, certainly $\dfrac{1-a}{a} \dfrac{1-b}{b} \geq 1$ if $a+b\leq\dfrac12.$

Here is another way to write the proof:

$\dfrac{1-a}a\dfrac{1-b}b\ge1\iff(1-a)(1-b)\ge ab\iff 1-a-b\ge0\iff1\ge a+b.$

Answer 2

$$\frac{1- a}{a}\times \frac{1- b}{b}> \frac{(\,a+ b\,)- a}{a}\times \frac{(\,a+ b\,)- b}{b}= 1$$

$$\because\,1> a+ b$$