South African Mathematics Olympiad Third Round

It is convenient to write $y = 1 - x$, so that $x + y = 1$. We show that the maximum of the resulting expression
$$\frac{1 - x^n - y^n}{xy^n + yx^n}$$
is attained when $x = y = \frac{1}{2}$. The value in this case is $2^n - 2$. We first rewrite the expression as follows:
$$\begin{aligned} \frac{1 - x^n - y^n}{xy^n + yx^n} &= \frac{x + y - x^n - y^n}{xy^n + yx^n} = \frac{x(1 - x^{n-1}) + y(1 - y^{n-1})}{xy(x^{n-1} + y^{n-1})} \\ &= \frac{x(1-x)(1+x+\cdots+x^{n-2}) + y(1-y)(1+y+\cdots+y^{n-2})}{xy(x^{n-1} + y^{n-1})} \\ &= \frac{xy(1+x+\cdots+x^{n-2}+1+y+\cdots+y^{n-2})}{xy(x^{n-1} + y^{n-1})} \\ &= \frac{1+1}{x^{n-1} + y^{n-1}} + \frac{x+y}{x^{n-1} + y^{n-1}} + \cdots + \frac{x^{n-2} + y^{n-2}}{x^{n-1} + y^{n-1}}. \end{aligned}$$

We will therefore be done if we can show that
$$\frac{x^a + y^a}{x^b + y^b}$$
attains its maximum for $x = y = \frac{1}{2}$ whenever $0 \le a < b$. This can be achieved in many ways, for example as follows: the general mean inequality yields
$$\left( \frac{x^a + y^a}{2} \right) \le \left( \frac{x^b + y^b}{2} \right)^{a/b}$$
and
$$\frac{1}{2} = \frac{x+y}{2} \le \left( \frac{x^b + y^b}{2} \right)^{1/b},$$

Using the same notation as in the previous solution, the binomial theorem gives us
$$\frac{1 - x^n - y^n}{xy^n + yx^n} = \frac{(x + y)^n - x^n - y^n}{xy^n + yx^n} = \frac{\sum_{k=1}^{n-1} \binom{n}{k} x^k y^{n-k}}{xy(x^{n-1} + y^{n-1})} = \frac{\sum_{k=1}^{n-1} \binom{n}{k} x^{k-1} y^{n-k-1}}{x^{n-1} + y^{n-1}}.$$
We group terms pairwise ($k$ and $n-k$ forming a pair):
$$\frac{1 - x^n - y^n}{xy^n + yx^n} = \sum_{k=1}^{\lfloor(n-1)/2\rfloor} \binom{n}{k} \frac{x^{k-1}y^{n-k-1} + x^{n-k-1}y^{k-1}}{x^{n-1} + y^{n-1}} + \begin{cases} \binom{n}{n/2} \frac{x^{n/2-1}y^{n/2-1}}{x^{n-1}+y^{n-1}} & n \text{ even,} \\ 0 & \text{otherwise.} \end{cases}$$
Now we only need to show that
$$\frac{x^{k-1}y^{n-k-1} + x^{n-k-1}y^{k-1}}{x^{n-1} + y^{n-1}} = \frac{(xy)^{k-1}(x^{n-2k} + y^{n-2k})}{x^{n-1} + y^{n-1}}$$
attains its maximum when $x = y = \frac{1}{2}$. Note that the potential extra term for even $n$ is also of this form, up to a factor 2:
$$\frac{x^{n/2-1}y^{n/2-1}}{x^{n-1} + y^{n-1}} = \frac{1}{2} \cdot \frac{(xy)^{n/2-1}(x^0 + y^0)}{x^{n-1} + y^{n-1}}.$$
Since $xy \le \left(\frac{x+y}{2}\right)^2 = \frac{1}{4}$ by the inequality between the arithmetic and geometric mean (with equality for $x = y = \frac{1}{2}$), this can again be achieved by showing that
$$\frac{x^a + y^a}{x^b + y^b}$$
attains its maximum for $x = y = \frac{1}{2}$ whenever $0 \le a < b$, as in the first solution.