<s>[INST] Prove or disprove: If $a+b \leq \frac{1}{2}$, then $\frac{(1-a)(1-b)}{ab} \geq 1$ for positive $a,b$ 
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?
 [/INST] $$\frac{1- a}{a}\times \frac{1- b}{b}> \frac{(\,a+ b\,)- a}{a}\times \frac{(\,a+ b\,)- b}{b}= 1$$
$$\because\,1> a+ b$$
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