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$\begin{aligned} f(x) & = \frac {(5+x)(2+x}{1+x} \\ & = \frac {x^2+7x+10}{x+1} \\ & = \frac {x^2+7x+10}{x+1} \\ & = \frac {(x+1)(x+6)+4}{x+1} \\ & = x + 6 + \frac 4{x+1} \end{aligned}$

Using AM-GM inequality on the following,

$\begin{aligned} {\color{#3D99F6}(x + 1)} + \frac 4{\color{#3D99F6}x+1} \ge 2 \sqrt 4 = 4 \\ \implies f(x) = x + 6 + \frac 4{x+1} \ge 4+5 = \boxed{9} \end{aligned}$

Equality occurs when $x+1 = \dfrac 4{x+1}$ $\implies (x+1)^2 = 4$ $\implies x = 1$ for $x>0$ .

[1] By graphing the curve.
OR
$[2] Let~~ f(x)=\dfrac{(5+x)*(2+x)}{1+x}.\\ The~the ~behavior~ of this~ type~ of~function:- \\ Its~range~is~-\infty \leq f(x) \leq +\infty. \\ Before~asymptote ~ it ~is ~like ~\pm~ a +\frac b c,~after~it~ is~ like ~\mp~e+\frac d c.\\ \therefore~it~ has~ only~one~local ~minimum~and~one~local~maximum. One ~extremum~on ~each~side.\\ So~for~our~function,~after~x=-1,~ there ~is~a~minimum. \\ f(.001)=9.997,~~~~~f(.002)=9.994,~~~~~f(.003)=9.991,~~~~decreasing.\\ f(.01)=9.970,~~~~~~~f(.02)=9.941,~~~~~~f(.03)=9.913,~~~~~decreasing.\\ f(.1)=9.997,~~~~~~~~~f(.2)=9.994,~~~~~~~f(.3)=9.991,~~~~~decreasing.\\ f(1)=9,~~~~~~~~~~~~~f(2)=9.333,~~~~~increasing.\\ f(9.000001)=15.4000,~~~~~~~~~f(8.999999)=15.399999.\\ So~ 9.000000 ~~is ~the~minimum.$

Let $y$ denote the expression given. Multiple $x+1$ on both sides of the equation. Obtain the quadratic in $x$ . Now since $x$ is real, the determinant has to be non-negative. This gives the condition on $y$ .