Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$. Describe when we have equality.
I tried simplifying the equation but got stuck. Can someone help me simplify this inequality into one that's easier to solve the actual problem above? I don't want/need the answer to the actual problem, but just need help simplifying the inequality. (A hint would be fine.) Any help would be appreciated! :)
Hint
$$\frac{1}{x} - 3 + 2\sqrt{x} = \frac{2x^{3/2} - 3x + 1}{x} = \frac{2t^3 - 3t^2 + 1}{t^2} (x = t^2)$$
Let $f(x)=\frac{1}{x}-3+2\sqrt{x}$
Then its derivative is defined and positive real numbers and equals $f'(x)=\frac{x^2-\sqrt{x}}{x^2 \sqrt{x}}$
You can easily show that $f'&lt;0$ if $x&lt;1$ and $f' \ge 0$ if $x\ge 1$.
Then $f$ reaches a minimum at $1$, which is $0$.
By considering the limits of $f$ at $0$ and $+\infty$, you can conclude.