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The minimum possible value of the largest of $ab$, $1-a-b+ab$, $a+b-2ab$ provided $0\leq a\leq b\leq 1$ is:

$1/3,\quad 4/9,\quad 5/9,\quad 1/9,\quad\text{ or }\quad \text{NONE}$

My approach :

I worked out that $\,1-a-b+ab\,$ will be the largest. Now $\,1-a-b+ab = (1-a)(1-b)\,$.

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    (1-a-b+ab)-(ab)=1-a-b which is not necessarily positive for the given conditions(take a=0.6,b=0.7).2012-06-27
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    Some trolls are downvoting without any explanation. I'd rather delete my answer...2012-06-27
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    Well each of the expressions is positive (since e.g. $a\geq ab$) so the minimum is bounded below by zero. Try some special cases ($a=b=\frac 1 2 , \text { or } a=b=\frac 1 3\text { or } a = \frac 1 3, b=\frac 2 3$ (taking thirds because of the answers suggested) and see what emerges.2012-06-27
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    Can you justify why it'll be a = b = 1/3 or a = 1/3, b = 2/3 please sir @MarkBennet2012-06-27
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    The sum $1 = ab + (1 - a - b + ab) + (a + b - 2ab)$ is at most $3$ times the maximum, with equality only where all are equal, which can't ever happen for real $a$ and $b$. So the largest of the three will always be greater than $1/3$..2012-06-27
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    @Bazinga - I wasn't suggesting it would be one of those values. But sometimes using easy values restricts the answers you are looking for and helps you to understand how the expressions you have been given behave. It is a potential way of making easy progress which does not take much time - which is why I suggested it. Multiplying thirds gives you ninths, which suggests using thirds as test values.2012-06-27
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    Where does this problem come from? The answer is surprisingly hard given that it is a multiple choice question.2012-06-27
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    As it turns out, each of the three gets a turn at being the largest. Consider (1) $a=b=0$, (2) $a=b=1$, and (3) $a=0,b=1$.2012-06-27

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Let $M(a,b)=\max\{ab,1-a-b+ab,a+b-2ab\}$, and observe that $$1=ab+(1-a-b+ab)+(a+b-2ab)\leq 3M(a,b),$$ so $$\frac{1}{3}\leq M(a,b)$$ for all such $a,b$. Hence, $\frac{1}{9}$ is ruled out. Can a value of $\frac{1}{3}$ actually be achieved? Well (as Cocopuffs pointed out), it can only be achieved if $\frac{1}{3}=ab=1-a-b+ab=a+b-2ab$, so let's see if that can happen.

Now, $ab=1-a-b+ab$ iff $1-a-b=0$ iff $a+b=1$. Hence, we'd need $a+b=1$ and $ab=\frac{1}{3},$ but this system yields $3a^2-3a+1=0$, which quadratic has a negative discriminant, and so no real solutions. Hence, there are no such real $a,b$, so $\frac{1}{3}$ will never be achieved, and we can rule it out as an answer.

If you try the special cases suggested by Mark Bennet, you'll see that it is possible for $M(a,b)=\frac{4}{9}$, so $\frac{5}{9}$ is ruled out as an answer. The only remaining question is: can we do better than $\frac{4}{9}$? Turns out that's the best we can do.

I really can't see a nice precalculus way to show that the answer is, in fact, $\frac{4}{9}$. Through methods of calculus, if we break $M(a,b)$ down piecewise, we can determine that it is minimized along the curves where the surface has a "fold", and then find that minimum explicitly. (See these W|A graphs for the idea. You'll want to click "Show contour lines" on the 3D plot.)