Curiosity on function maxima I was recently working with an equation of the form:
$$
\frac{\sqrt{x}}{a+bx+c\sqrt{x}}
$$
And I realized that the maxima (only considering positive real numbers) would always be at the point where:
$$
x=\frac ab 
$$
This is straightforward to prove by finding where the first derivative equals 0. Given this 'easy' result, I tried to find the logic behind it, which should probably be something easy, but I do not find it (I'm evidently no expert in mathematics, just curious).
My question is, should it be evident that the function has a maxima at that point without having to calculate the derivative? In the case it should, could someone explain me the reasoning behind it?
Thank you in advance.
Kind regards,
J.
 A: Write it as:$$
\frac{1}{b\sqrt{x}+\cfrac{a}{\sqrt{x}}+c}
$$
Then by AM-GM:
$$
b\sqrt{x}+\cfrac{a}{\sqrt{x}} \ge 2 \sqrt{ab}
$$
Also by AM-GM, equality holds when $b\sqrt{x}=\cfrac{a}{\sqrt{x}} \iff x = \cfrac{a}{b}\,$, which thus gives the minimum of the denominator, which in turn gives the maximum of the fraction.

[ EDIT ]   The above interpretes the "only considering positive real numbers" stated condition to mean that $a,b,c$ and $x$ are strictly positive numbers.
A: The reciprocal function is
$$\frac a{\sqrt x}+b\sqrt x+c$$ and the position of its extrema is independent of $c$.
We can factor out $b$ and get
$$b\left(\frac ab\frac 1{\sqrt x}+\sqrt x\right)+c,$$ which shows that the position can only depend on $\dfrac ab$.
The term $\dfrac a{\sqrt x}$ is decreasing and $b\sqrt x$ is increasing, the extremum is achieved when their slopes are opposite, which occurs when 
$$\frac a{2x\sqrt x}=\frac b{2\sqrt x}.$$
A: the first derivative is given by $$f'(x)=1/2\,{\frac {-bx+a}{ \left( a+bx+c\sqrt {x} \right) ^{2}\sqrt {x}}}$$ the searched extrema ( if they exist) are located at $$x=\frac{a}{b}$$
