Prove that the maximum of the following expression occurs when all $\dfrac{x_{i+1}}{x_i}$ are equal for $0\leq i< n.$

Question

Suppose $0< a=x_0 < x_1 < \dots < x_n =b.$ Show that the maximum of the following expression $$\dfrac{x_0x_1...x_n}{(x_0+x_1)(x_1+x_2)\dots (x_{n-1}+x_n)}$$ occurs only when all $\dfrac{x_{i+1}}{x_i}$ are equal for $0\leq i < n.$

I thought differentiating this function somehow. I thought Jensen's inequality might be useful for this. I'm pretty sure I need to find an upper bound for this expression using Jensen's inequality and then make it obvious that this only occurs when all the $\dfrac{x_{i+1}}{x_i}$'s are equal.

Also, I think I can use partial fraction decomposition for this problem.

Answer 1

Since the expresion $$\dfrac{x_0x_1...x_n}{(x_0+x_1)(x_1+x_2)\dots (x_{n-1}+x_n)}$$ is positive, it is maximal iff this is minimal $$\dfrac{(x_0+x_1)(x_1+x_2)\dots (x_{n-1}+x_n)}{x_0x_1...x_n}$$

which the same as $$I=(1+ {x_1\over x_0})(1+ {x_2\over x_1})...(1+ {x_n\over x_{n-1}})$$

Now by $x+y\geq 2\sqrt{xy}$ we have $$I \geq 2^n\sqrt{x_n\over x_0}= 2^n\sqrt{b\over a}$$

and the equality is achived if ...