Does $1-ab\geq (1-a)(1-b)$ hold, for $a,b\in [0,1]$?

Question

Does this inequality hold, for any $a,b\in[0,1]$

$$1-ab\geq (1-a)(1-b)?$$

I'm don't have idea to conclude $1-ab\geq 1-a-b+ab = (1-a)(1-b)$.

Anyone can prove (or disprove) it?

Answer 1

No. Take $a=0$ and $b=1$ to disprove it.

Addendum:

Now that you changed the inequality, it follows from AM-GM:

$a+b\ge2\sqrt a\sqrt b\ge 2ab\iff 1-ab\ge1+ab-a-b=(1-a)(1-b)$.

Answer 2

Hint:

as $1-a\ge 0,1-b\ge 0$ $$1-ab-(1-a)(1-b)=a(1-b)+b(1-a)\ge 0$$

Answer 3

I approached this question geometrically.

Desmos for fun

Start with a square of side length 1. Divide the square along one side in the ratio $a:(1-a)$, and the other side $b:(1-b)$. By considering the relevant areas given by $(1-ab)$ and $(1-a)(1-b)$, it should be obvious the inequality is true.

Answer 4

$$\text{RHS} - \text{LHS} = 1+ab -a-b - (1-ab) = 2ab-a-b ≤ 2ab-a^2 - b^2 = -(a-b)^2 ≤ 0. $$ Thus $\text{LHS} \ge \text{RHS}$.

Answer 5

Define $m=2a-1$ i.e. $m\in [-1 , 1]$ and $n=b-\frac{1}{2}$ i.e. $n\in [-\frac{1}{2} , \frac{1}{2}]$. Is the following true?

$$ \begin{align} -\frac{1}{2}&\leq mn\leq \frac{1}{2}\\ -ab&\leq (1-a)(1-b)\leq 1-ab \end{align} $$

Answer 6

$1-ab \ge 1-a \ge (1-a)(1-b).\blacksquare$

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Update: I found this is equivalent to that of @Albus Dumbledore and @Tony Ip.