AustriaMO2011

Since we have $x^2+y^2=1$, it definitely follows that $|x|\le 1$ and $|y|\le 1$ hold. Defining a function $g_k(x):=x-x^{2k+1}$, the signs of $x$ and $g_k(x)$ are therefore equal, and we have $g_k(-x)=-g_k(x)$. The given function can be expressed as $f_k(x,y)=g_k(x)+g_k(y)$, and we certainly have $f_k(x,y)\le f_k(|x|,|y|)$. Since $x^2+y^2=1$ implies $|x{|}^2+|y{|}^2=1$, we can assume that $x,y\ge 0$ holds, which allows us to apply the means inequality. For the quadratic mean, we have $$m_2(x,y)=\sqrt{\frac{x^2+y^2}{2}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2},$$ and from $$x+y=2m_1(x,y)\le 2m_2(x,y)\le 2m_{2k+1}(x,y)$$ we obtain $$-(x^{2k+1}+y^{2k+1})=-2m_{2k+1}^2(x,y)\le -2m_2^2(x,y).$$ We therefore have both $x+y\le \sqrt{2}$ and $$-(x^{2k+1}+y^{2k+1})\le \frac{2}{{\sqrt{2}}^{2k+1}}=\frac{\sqrt{2}}{2^k},$$ which imply $$f_k(x,y)\le \frac{2^k-1}{2^k}\sqrt{2}$$ with equality holding for $x=y=\frac{\sqrt{2}}{2}$. qed