An algebra problem by William Isoroku

When $x=2$ the expression (not equation as there is no equality) is equal to 3.

But $\frac {(x+4)(x+1)}{x}=\frac {x^{2}+5x+4}{x}=x+5+\frac {4}{x}=(\sqrt {x}-\frac {2}{\sqrt {x}})^{2}+9 \geq 9$ . The result follows.

Alternatively AM-GM on $x, \frac {4}{x}$ can be used for the last step.

$By\quad AM-GM,\\ x+4\ge 2\sqrt { 4x } \\ x+1\ge 2\sqrt { x } \\ Therefore,\quad \left( x+4 \right) \left( x+1 \right) \ge 8x,\\ Dividing\quad both\quad sides\quad by\quad x,\quad we\quad get\quad \left( x+4 \right) \left( x+1 \right) /x\ge 8,\\ Taking\quad square\quad roots\quad of\quad both\quad sides\quad gives\quad the\quad desired\quad solution.\quad \\ Although,\quad you\quad don't\quad actually\quad get\quad 3$

simplify inside the root and differentiate it put the derivative =0 and get value of x finally substitue and get the answer

I used the first derivative and as the domain is x>0 the only sign change occurs at x=2 from - to + . So f(2)=fmin.