Let $x$ be a positive real. Find the maximum possible value of the expression $y = \frac{x^{2} + 2 - \sqrt{x^{4} + 4}}{x}$

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Solution

$y = \frac{x^{2} + 2 - \sqrt{x^{4} + 4}}{x} \frac{d y}{d x} = \frac{x (2 x - \frac{4 x^{3}}{2 \sqrt{x^{4} + 4}}) - {(x}^{2} + 2 - \sqrt{x^{4} + 4})}{x^{2}} \frac{d y}{d x} = 0 x (2 x - \frac{4 x^{3}}{2 \sqrt{x^{4} + 4}}) = {(x}^{2} + 2 - \sqrt{x^{4} + 4}) x^{2} - 2 = \frac{2 x^{4} - 8}{2 \sqrt{x^{4} + 4}} x^{2} - 2 = \frac{x^{4} - 4}{\sqrt{x^{4} + 4}} x^{2} - 2 = \frac{{(x}^{2} + 2) {(x}^{2} - 2)}{\sqrt{x^{4} + 4}} x^{4} + 4 = x^{4} + 4 + 4 x^{2} x = 0 y a t 0 = \frac{{(x}^{2} + 2) - \sqrt{x^{4} + 4}}{x} u s i n g L H R = 2 x - \frac{4 x^{3}}{2 \sqrt{x^{4} + 4}} = 0$

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