It seems that max value will occur at $a=1$.

Question

For $a \in\{1,2,\ldots,n-1\}$, I need to maximize the following: $$\!\left(a\sqrt{\frac{a}{a+1}}+\sqrt{\frac{a(n-a)}{(a+1)(n-a+1)}}\right) \left((n-a)\sqrt{\frac{n-a}{n-a+1}}+\sqrt{\frac{a(n-a)}{(a+1)(n-a+1)}}\right) $$ It seems that max value will occur at $a=1$. Since above is a symmetric function in variable $a,n-a$, max value will also occur at $a=n-1.$

I am not able to prove that maximum will occur at $a=1$. How to prove it?

Answer 1

Your assumption that the maximum will occur when $a=1$ is false so you can't prove it. Look at the this wolfram alpha plot (eg. I've set the n=100 so the plot is not 3D).

Answer 2

HINT.-The expression $$\!\left(a\sqrt{\frac{a}{a+1}}+\sqrt{\frac{a(n-a)}{(a+1)(n-a+1)}}\right) \left((n-a)\sqrt{\frac{n-a}{n-a+1}}+\sqrt{\frac{a(n-a)}{(a+1)(n-a+1)}}\right)$$ is factorized as $$\sqrt{\frac{a(n-a)}{(a+1)(n-a+1)}}\left(a+\sqrt{\frac{n-a}{n-a+1}}\right)\left(n-a+\sqrt\frac{a}{a+1}\right)$$

You have to consider the function $$F_n(x)=\sqrt{\frac{x(n-x)}{(x+1)(n-x+1)}}\left(x+\sqrt{\frac{n-x}{n-x+1}}\right)\left(n-x+\sqrt\frac{x}{x+1}\right)$$ where you restraint $x=1,2,3\cdots,n-1$ and to verify that the maximum for these integer values is reach with $x=1$.

Taking derivative $(F_n(x))'=0$ is too hard to solve. Graphically you can verify that the maximum (for $x\gt0$) is done for $x\approx 1$ and $F_n(1)\gt F_n(2)\gt\cdots F_n(n-1)\gt F_n(n)=0$ for many values. However, for instance, for $n=8$ you have $$\text{Max }F_8(1.493)=46.85\\F_8(1)\approx 45.55\\F_8(2)\approx45.99\\F_8(3)\approx41.33$$ So the maximum for the integer values $1,2,3,4,5,6,7$ is reach with $a=2$ instead of with $a=1$. The statement is not true.