Minimo di $f(x)=x+\dfrac{9}{x}$ per $x>0$

Text

Find the minimum value of $f(x)=x+\frac{9}{x},\qquad x>0,$ and the point where it is attained.

Solution

Apply the arithmetic–geometric mean inequality to the two positive terms $x$ and $\dfrac9x$: $\frac{x+\frac9x}{2}\ge\sqrt{x\cdot\frac9x}=\sqrt9=3\implies x+\frac9x\ge 6.$ Equality (hence the minimum) occurs when $x=\dfrac9x$, i.e. $x^2=9$ and, since $x>0$, $x=3$. $\boxed{f_{\min}=6\ \text{ at } x=3.}$