Maths Olympiad Prep

Library / /718 of 860

Algebra Difficulty 5.4 AIME, harder Find the answer

Let xx be a positive real number. Find the maximum possible value of x2+2x4+4x\frac{x^{2}+2-\sqrt{x^{4}+4}}{x}

A number or a short expression. Spacing and $ signs are ignored.

Solution

Rationalizing the numerator, we get x2+2x4+4xx2+2+x4+4x2+2+x4+4=(x2+2)2(x4+4)x(x2+2+x4+4)=4x2x(x2+2+x4+4)=41x(x2+2+x4+4)=4x+2x+x2+4x2\begin{aligned} \frac{x^{2}+2-\sqrt{x^{4}+4}}{x} \cdot \frac{x^{2}+2+\sqrt{x^{4}+4}}{x^{2}+2+\sqrt{x^{4}+4}} & =\frac{\left(x^{2}+2\right)^{2}-\left(x^{4}+4\right)}{x\left(x^{2}+2+\sqrt{x^{4}+4}\right)} \\ & =\frac{4 x^{2}}{x\left(x^{2}+2+\sqrt{x^{4}+4}\right)} \\ & =\frac{4}{\frac{1}{x}\left(x^{2}+2+\sqrt{x^{4}+4}\right)} \\ & =\frac{4}{x+\frac{2}{x}+\sqrt{x^{2}+\frac{4}{x^{2}}}} \end{aligned} Since we wish to maximize this quantity, we wish to minimize the denominator. By AM-GM, x+2x22x+\frac{2}{x} \geq 2 \sqrt{2} and x2+4x24x^{2}+\frac{4}{x^{2}} \geq 4, so that the denominator is at least 22+22 \sqrt{2}+2. Therefore, x2+2x4+4x422+2=222,\frac{x^{2}+2-\sqrt{x^{4}+4}}{x} \leq \frac{4}{2 \sqrt{2}+2}=2 \sqrt{2}-2, with equality when x=2x=\sqrt{2}.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.