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$x$ and $y$ are positive real numbers.

How can we find the minimum value of the sum $(16/x)+(108/y)+(xy)$ ?

It seems to be a easy question but I could not even decide where to start...

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    I would first try it without the restriction that $x$ and $y$ must be positive. Write your sum as a function of two variables: $$ f(x, y) = (16/x)+(108/y)+(xy), $$ find its gradient, equate it to the zero vector, and see which values of $x$ and $y$, if any, satisfy the two equations. – 2017-01-17
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    Did you try setting the gradient equal to $0$? – 2017-01-17
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    this question is located in a high school maths book. So I looked for more simple solutions. Thanks. – 2017-01-17

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We have, by the AM-GM inequality, $$ \frac{16}x + \frac{108}{y} + xy\geq 3\sqrt[3]{\frac{16}x \cdot\frac{108}{y} \cdot xy} = \boxed{36} $$ with equality iff all the three terms $\frac{16}x,\frac{108}{y}$ and $xy$ are equal. Thus, in order to find this minimum, we want to solve the simultaneous equations $$\begin{cases} \frac{16}{x} = xy\\ \frac{108}{y} = xy\end{cases}\\ \begin{cases} 16 = x^2y\\ 108 = xy^2\end{cases} $$ Squaring the top equation, and then dividing by the bottom one gives $$ \frac{256}{108} = \frac{x^4y^2}{xy^2}\\ \frac{64}{27} = x^3\\ \boxed{\frac{4}{3} = x} $$ which then gives us $16 = \left(\frac{4}{3}\right)^2\! y$, or $\boxed{y = 9}$.

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Since $16/x$ and $108/y$ blow up when $x$ or $y$ go to zero and $xy$ blows up as $x$ or $y$ go to infinity (when neither are zero) you know the minimum is somewhere in the interior of the first quadrant. So you need to find the critical point(s) where $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y} = 0.$$ Your minimum will be at a critical point.