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 $$\frac{1-x^n-y^n}{x{y}^n+y{x}^n}$$ is attained when $x=y=\frac{1}{2}$. The value in this case is $2^n-2$. We first rewrite the expression as follows: $$\begin{array}{rl}\frac{1-x^n-y^n}{x{y}^n+y{x}^n}& =\frac{x+y-x^n-y^n}{x{y}^n+y{x}^n}=\frac{x(1-x^{n-1})+y(1-y^{n-1})}{xy(x^{n-1}+y^{n-1})}\\ & =\frac{x(1-x)(1+x+\cdots +x^{n-2})+y(1-y)(1+y+\cdots +y^{n-2})}{xy(x^{n-1}+y^{n-1})}\\ & =\frac{xy(1+x+\cdots +x^{n-2}+1+y+\cdots +y^{n-2})}{xy(x^{n-1}+y^{n-1})}\\ & =\frac{1+1}{x^{n-1}+y^{n-1}}+\frac{x+y}{x^{n-1}+y^{n-1}}+\cdots +\frac{x^{n-2}+y^{n-2}}{x^{n-1}+y^{n-1}}.\end{array}$$

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 $$\frac{1}{2}=\frac{x+y}{2}\le {\left(\frac{x^b+y^b}{2}\right)}^{1/b},$$

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Alternative solution.

Using the same notation as in the previous solution, the binomial theorem gives us $$\frac{1-x^n-y^n}{x{y}^n+y{x}^n}=\frac{(x+y{)}^n-x^n-y^n}{x{y}^n+y{x}^n}=\frac{{\sum }_{k=1}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{k}\right)x^k{y}^{n-k}}{xy(x^{n-1}+y^{n-1})}=\frac{{\sum }_{k=1}^{n-1}\left(\genfrac{}{}{0ex}{}{n}{k}\right)x^{k-1}{y}^{n-k-1}}{x^{n-1}+y^{n-1}}.$$ We group terms pairwise ($k$ and $n-k$ forming a pair): $$\frac{1-x^n-y^n}{x{y}^n+y{x}^n}=\sum _{k=1}^{\lfloor (n-1)/2\rfloor }\left(\genfrac{}{}{0ex}{}{n}{k}\right)\frac{x^{k-1}{y}^{n-k-1}+x^{n-k-1}{y}^{k-1}}{x^{n-1}+y^{n-1}}+\{\begin{array}{ll}\left(\genfrac{}{}{0ex}{}{n}{n/2}\right)\frac{x^{n/2-1}{y}^{n/2-1}}{x^{n-1}+y^{n-1}}& n\text{ even,}\\ 0& \text{otherwise.}\end{array}$$ 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.