# Arhimede International Mathematics Competition 2008

Problem
Let $x,y$ be reals s.t. $x^2y^2\leq1$ and $n$ a natural number. Prove that:
$$(x^n+y)^2+y^2\geq\dfrac{1}{n+2}(x^2+y^2)^n$$
Võ Quốc Bá Cẩn’s solution
Using the Cauchy-Schwarz inequality, we have $(x^n+y)^2+\frac{2}{n+2}y^2 \ge \frac{(x^n+y-y)^2}{1+\frac{n+2}{2}}=\frac{2}{n+4}x^{2n}.$ Therefore, \begin{aligned} (x^n+y)^2+y^2&=\left[(x^n+y)^2+\frac{2}{n+2}y^2\right] +\frac{n}{n+2}y^2 \ge \frac{2}{n+4}x^{2n}+\frac{n}{n+2}y^2\\ &\ge \frac{2}{n+4}x^{2n}+\frac{n}{n+2} y^2(x^2+y^2)^{n-1}. \end{aligned} It suffices to prove that $\frac{2}{n+4}x^{2n}+\frac{n}{n+2}y^2(x^2+y^2)^{n-1} \ge \frac{1}{n+2}(x^2+y^2)^2,$ or $\frac{2n+4}{n+4}x^{2n} \ge (x^2+y^2)^{n-1} \big[x^2-(n-1)y^2\big].$ If $x^2 \le (n-1)y^2,$ then the inequality is obviously true. Otherwise, using the AM-GM inequality, we have \begin{aligned} (x^2+y^2)^{n-1}\big[x^2-(n-1)y^2\big] &\le \left[\frac{(n-1)(x^2+y^2)+x^2-(n-1)y^2}{n}\right]^n\\ &=x^{2n} \le \frac{2n+4}{n+4} x^{2n}.\end{aligned} This ends our proof.