How to show that in function $f(x)=\frac{x}{(x+b)^2}$ maximum value of $f(x)$ is when $x = b$?

Question

How to show that in function $f(x)=\frac{x}{(x+b)^2}$ maximum value of $f(x)$ is when $$x = b$$?
EDIT: I saw solution using derivatives but I don't know nothing about derivatives. Can it be done without derivatives?
$b, x > 0$

Answer 1

Hint:

You want to prove that : $$ \frac{x}{(x+b)^2}\le \frac{1}{4b} $$ For $b>0$ becomes: $4bx\le (x+b)^2$ and it is easy to prove that it is true for all $x$.

The case $b<0$ is for you .

Answer 2

It was shown that for $f(x)=\frac{x}{(x+b)^2}$: $$f'(x)=\frac{(x+b)^2-2x(x+b)}{(x+b)^4}=\frac{(x+b)-2x}{(x+b)^3}=\frac{b-x}{(x+b)^3}$$ Thus, $f'(x)=0$ when $x=b$. To see $x=b$ is minimum or maximum, $$f''(x)=\frac{(x+b)^3(-1)-3(b-x)(x+b)^2}{(x+b)^6}=\frac{-(x+b)-3(b-x)}{(x+b)^4}=\frac{2(x-2b)}{(x+b)^4}$$ When $x=b$, $f''(x)\lt0$, hence, $x=b$ point is a maximum.

Answer 3

Hint:  given that $\,b,x \gt 0\,$, it follows from AM-GM that the following inequality holds, with equality iff $\sqrt{x}=\dfrac{b}{\sqrt{x}} \iff x = b$:

$$\require{cancel} \frac{x}{(x+b)^2} \;=\; \frac{1}{\left(\sqrt{x}+\cfrac{b}{\sqrt{x}}\right)^2} \;\le\; \frac{1}{\left(2 \sqrt{\cancel{\sqrt{x}}\cfrac{b}{\cancel{\sqrt{x}}}}\,\right)^2} \;=\; \frac{1}{4b} $$