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

Question

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.

Answer 1

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

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

Answer 2

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.

Answer 3

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 $$0<c\le 1$$ for $$0\le x\le c$$ or $$c>1$$ and $$0\le x\le 1$$