let x be positive real number, find max possible value of the expression $$y = \frac{x^2 + 2 - \sqrt{x^4 + 4}}{x}$$
it can be found by differentiating, but is there no other way of finding it, like using AM $\geq$ GM. or any other method.
i tried $$y = x + \frac{2}{x} - \sqrt{x^2 + \frac{4}{x^2}}$$ but it gives nothing
On
You are on the right track!
To begin with, suppose that $x > 0$. Then we can rearrange the proposed expression as \begin{align*} f(x) = x + \frac{2}{x} - \sqrt{x^{2} + \frac{4}{x^{2}}} = x + \frac{2}{x} - \sqrt{\left(x + \frac{2}{x}\right)^{2} - 4} \end{align*}
If we make the substitution $u = x + \dfrac{2}{x}$, one gets: \begin{align*} f(u) = u - \sqrt{u^{2} - 4} \end{align*}
Now you can apply the derivative method. Analogous approach applies to the case where $x < 0$.
Can you take it from here?
$$y(x) = x + \frac{2}{x} - \sqrt{x^2 + \frac{4}{x^2}} = x + \frac{2}{x} -\sqrt{(x + \frac{2}{x})^2-4}$$ $$t=x+\frac{2}{x}$$ $$y(t)=t-\sqrt{t^2-4} =\frac{4}{t+\sqrt{t^2-4}}$$ which is clearly monotonicly decreasing.
so we need to find the minimal value of $x+\frac{2}{x}$ by AM-GM: $$\frac{x+\frac{2}{x}}{2}\geq \sqrt{2}$$