Maximum value of a symmetric function on a square In a paper of Frits Beukers "A note on the irrationality of $\zeta(2)$ and $\zeta(3)$", he says

It is a matter of straightforward computation to show that
$$\frac{y(1-y)x(1-x)}{1-xy} \le \left\{\frac{\sqrt{5}-1}{2}\right\}^5$$ for all $0\le x\le1, 0\le y \le 1$.

I noticed (using calculus) that if we let $x = y$,
$$
\frac{x^2(1-x)^2}{1-x^2} = \frac{x^2(1-x)}{1+x}
$$
has a maximum value of $\left\{\frac{\sqrt{5}-1}{2}\right\}^5$ on the interval $0\le x\le1$.
My question is why is the maximum of $\frac{y(1-y)x(1-x)}{1-xy}$ attained when $x=y$? Is it because of the symmetry in $x$ and $y$?.
Is it true in general that a symmetric function $f(x_1,x_2,\dots,x_n)$ attains its maximum on an $n$-dimensional cube when $x_1=x_2=\dots=x_n$?

$$\frac{y(1-y)x(1-x)}{1-xy} = \frac{xy(1-(x+y)+xy)}{1-xy} \le \frac{xy(1-2\sqrt{xy}+xy)}{1-xy}= \frac{xy(1-\sqrt{xy})}{1+\sqrt{xy}}$$
Denote $t = \sqrt{xy}$ then $0 \le t\le 1$.
The function $f(t) = \frac{t^2(1-t)}{1+t}$ reaches its maximum in $t \in [0,1]$ when $t = \frac{\sqrt{5}-1}{2}$ (it suffices to calculate $f'(t)$)
The equality occurs when $x = y$ (so that $x+y = 2\sqrt{xy}$) and $\sqrt{xy} = \frac{\sqrt{5}-1}{2}$, or $x = y = \frac{\sqrt{5}-1}{2}$
PS: about your last question concerning the relationship between the symmetry of $f(x_1,...,x_n)$ and the maximum of $f$, this relationship doesn't exists in general. And in this particular problem, the equality attains because $x+y = \sqrt{xy} \iff x = y$.
