How to find the maximum value of a quadratic equation?

Question

I have the expression $\displaystyle y = \frac{x^2+2-\sqrt{x^4+4}}x$. I want to find its maximum value when x is a positive real number. I proceeded like this, but don't know if the process is right.

Since it asked for the maximum value,the term inside the square root must be the least(as it is subtracted) and the least it can be is zero. Hence $ x^4+4=0$, thus $x^4=-4$. But now $x$ will be imaginary, but it must be real.

I also tried to use the formula $\displaystyle\frac {-b\pm\sqrt{b^2-4ac}}{2a}$, but I don't know how to use it in this equation.

Answer 1

First of all, we need $x>0$

Rationalizing the numerator $$y=\dfrac{4x}{x^2+2+\sqrt{x^4+4}}=\dfrac4{x+\dfrac2x+\sqrt{\left(x+\dfrac2x\right)^2-4}}$$

Now using AM-GM inequality, $$\dfrac{x+\dfrac2x}2\ge\sqrt{x\cdot\dfrac2x}=?$$

Answer 2

By AM-GM

$$y=x+\frac{2}{x}-\sqrt{\left(x+\frac{2}{x}\right)^2-4}\leq2\sqrt{x\cdot\frac{2}{x}}-\sqrt{\left(2\sqrt{x\cdot\frac{2}{x}}\right)^2-4}=2\sqrt2-2$$ because $f(t)=t-\sqrt{t^2-4}$ is a decreasing function.

Indeed, $f'(t)=1-\frac{t}{\sqrt{t^2-4}}<0$.

The equality occurs for $x=\frac{2}{x},$ which says that $2\sqrt2-2$ is the answer.

Done!