Vietnam Math Out Standing Student 2012-13

Vietnam Math Outstanding Student 2012-2013

Problem 01: Solve the equation \(x^{4n}+\sqrt{x^{2n}+2012}=2012\), for all \(n\in\mathbb{N}\).

Problem 02: It is given the sequence \((U_n)\) determined by \(U_1=3\) and \(U_{n+1}=\frac13\left(2U_n+\frac3{U_n^2}\right)\), for all \(n\in\mathbb{N}\). Find the limit \(\lim_{n\rightarrow\infty}U_n\).

Problem 03: Given three non-negative real numbers \(x,y,z\), prove that \(\frac1x+\frac1y+\frac1z\geq\frac{36}{9+x^2y^2+y^2z^2+z^2x^2}\).

Problem 04: Find the positive roots of the equation \(\sqrt{x+2\sqrt3}=\sqrt y+\sqrt z\).

Vietnam Math Outstanding Student 2012-2013

Problem 01: Solve the equation \(x^{4n}+\sqrt{x^{2n}+2012}=2012\), for all \(n\in\mathbb{N}\).

Solution: Let \(t=x^{2n}\). Then the equation becomes \(t^2+\sqrt{t+2012}=2012\).

We have \(t^2=2012-\sqrt{t+2012}\), so adding \(t+\frac14\) to both sides gives \((t+\frac12)^2=t+2012-\sqrt{t+2012}+\frac14=(\sqrt{t+2012}-\frac12)^2\).

Since \(t=x^{2n}\geq0\), we get \(t+\frac12=\sqrt{t+2012}-\frac12\), hence \(t+1=\sqrt{t+2012}\).

Squaring both sides, \(t^2+t+1=t+2012\), so \(t^2-2011=0\).

Therefore, \(t=\sqrt{2011}\), and since \(t=x^{2n}\), we have \(x^{2n}=\sqrt{2011}\).