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Suppose two numbers $a$ and $b$ such that $a+b=1$.

How can I prove that the product is $ab\le \frac{1}{4}$ ?

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    Before your edit you wrote numbers $p$ and $1-p$,you were on the right track,to finish try to adjust the inequality so it's $\text{something}\leq 0$. – 2017-02-12

4 Answers 4

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Depends what you are supposed to use. One simple way is $$a+b=1 \Rightarrow a = 1-b.$$ Thus, $$ab = (1-b)b = b-b^2.$$ Therefore, the max of $ab$ is attained when $$\frac{d}{db} \left(b-b^2\right) = 1-2b = 0 \Rightarrow b =\frac{1}{2}.$$ Thus, $$\max\{ab\} = \left(1-\frac{1}{2}\right)\cdot\frac{1}{2} = \frac{1}{4}.$$

Edit: To be complete (as asked about in the comments), we can use the "first derivative test" to demonstrate that this is indeed a maximum. In particular, for $b = \frac{1}{2} - \epsilon$ ($\epsilon>0$), $$1 - 2b = 1-2\left(\frac{1}{2} - \epsilon\right) = \epsilon > 0,$$ and similarly for $b = \frac{1}{2} + \epsilon,$ $$1 - 2b = 1-2\left(\frac{1}{2} + \epsilon\right) = -\epsilon < 0.$$ Hence, $b = a = 1/2$ is indeed a maximum.

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    $\dfrac{d^2}{db^2}(b)=0$ so how can you say $b=\frac 12$ is maximum? – 2017-02-12
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    the "first derivative test" works... – 2017-02-12
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Another way. Let $a = \frac 12 + d $. Then $b =1-a =\frac 12 -d $. So $ab = (\frac 12 +d) (\frac 12 -d)=\frac 14 -d^2 \le \frac 14$

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Hint: Of course if one of $a$ and $b$ is negative, then their product is necessarily negative. So, suppose that $a,b \geq 0$. Note that $(\sqrt{a} - \sqrt{b})^2$ is non-negative.

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    Got it! Thanks a lot. I didn't see that option – 2017-02-12
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We need to prove that $$ab\leq\frac{1}{4}(a+b)^2$$ or $$(a-b)^2\geq0$$