Show that $\frac 1x \ge 3 - 2\sqrt{x}$ for all positive real numbers $x$.

Question

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! :)

Answer 1

Hint $$\frac{1}{x} - 3 + 2\sqrt{x} = \frac{2x^{3/2} - 3x + 1}{x} = \frac{2t^3 - 3t^2 + 1}{t^2} (x = t^2)$$

Answer 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 $f'<0$ if $x<1$ 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.

Answer 3

We need to prove that: $$2x\sqrt{x}-3x+1\geq0,$$ which is true by AM-GM: $$2x\sqrt{x}-3x+1\geq3\sqrt[3]{(x\sqrt{x})^2\cdot1}-3x=0.$$

Answer 4

Hint : $\quad$ You have to prove

$$\frac{1}{x} + \sqrt{x} + \sqrt{x} \geq 3$$