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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. triangle The question becomes finding the longest length of the red line. However, I have no idea how to proceed.

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3 Answers 3

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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.

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    $\begingroup$ Hi Michael thanks again for you kind support during the suspension time. If you are interested I've raised a related discussion HERE. Cheers $\endgroup$
    – user
    Oct 14, 2018 at 10:26
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    $\begingroup$ 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. $\endgroup$ Oct 14, 2018 at 12:38
  • $\begingroup$ I think that open discussion are the only way to find a compromise between different point of views. Thanks again. Bye $\endgroup$
    – user
    Oct 14, 2018 at 12:46
  • $\begingroup$ Nice approach but somewhat cheating to take $x=\frac{c}{c+1}t$ at the beginning. $\endgroup$
    – user328174
    Oct 15, 2018 at 4:44
  • $\begingroup$ @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 $1$. $\endgroup$ Oct 15, 2018 at 4:47
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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 $$0\le x\le 1$$

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

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

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