# Maximize the value of $\sqrt{x-x^2}+\sqrt{cx-x^2}$ without using calculus

Assume that $$c$$ is positive. How can we maximize the value of $$\sqrt{x-x^2}+\sqrt{cx-x^2}$$ with respect to $$x$$ without the use of calculus?

With calculus, we can easily find out that the max of the expression is when $$x=\frac{c}{c+1}$$.

My attempt to the question is consider the expression as the distance between points. Below is the figure. The question becomes finding the longest length of the red line. However, I have no idea how to proceed.

Let $$x=\frac{c}{c+1}t.$$

Thus, by AM-GM we obtain: $$\sqrt{x-x^2}+\sqrt{cx-x^2}=\frac{\sqrt{c}}{c+1}\sqrt{t(c+1-ct)}+\frac{\sqrt{c}}{c+1}\sqrt{ct(c+1-t)}\leq$$ $$\leq\frac{\sqrt{c}}{c+1}\left(\frac{t+c+1-ct}{2}+\frac{ct+c+1-t}{2}\right)=\sqrt{c}.$$ THe equality occurs for $$t=1$$, which says that we got a maximal value.

• Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers
– user
Oct 14, 2018 at 10:26
• Hi @gimusi . I really don't like discussions in Meta. In my opinion this a place, where a limited number of users, which are destroying this site, trying to get a legitimation to do it. Oct 14, 2018 at 12:38
• I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye
– user
Oct 14, 2018 at 12:46
• Nice approach but somewhat cheating to take $x=\frac{c}{c+1}t$ at the beginning. Oct 15, 2018 at 4:44
• @Wss Because we know that the equality occurs for $x=\frac{c}{c +1}$ and it would be much more better if it will happen for some variable is equal to $1$. Oct 15, 2018 at 4:47

Assuming $$\sqrt{x-x^2}+\sqrt{cx-x^2}\le \sqrt{c}$$ Squaring we get

$$2\sqrt{x-x^2}\sqrt{cx-x^2}\le 2x^2-x-cx+c$$ squaring again and factorizing we get $$(cx-c+x)^2\geq 0$$ which is true. Remark: We can only square the inequality if $$2x^2-x(c+1)+c\geq 0$$ this is true if $$0 for $$0\le x\le c$$ or $$c>1$$ and $$0\le x\le 1$$

• @Sonnhard I think the right side of your second inequality can be negative. At least, you need to prove that it's not negative. Oct 14, 2018 at 9:27
• This is true, the condition is complicated. Oct 14, 2018 at 9:38
• @Sonnhard But $c^2-6c+1$ can be positive, which says that the right side can be negative. No? Oct 14, 2018 at 9:41
• Yes with the inequalities $$x-x^2\geq 0$$ and$$cx-x^2\geq 0$$ it simplifies to the given above Oct 14, 2018 at 9:53
• @Sonnhard But how did you got it? By the way, there is a very nice proof that $2x^2-(c+1)x+1>0$. Try to find it. Oct 14, 2018 at 9:59

Using the Cauchy-Schwartz inequality

$$\left(\sqrt{x-x^2}+\sqrt{c x-x^2}\right)^2\le \left(x+1-x\right)\left(x+c-x\right) = c$$

then

$$\sqrt{x-x^2}+\sqrt{c x-x^2}\le\sqrt{c}$$