maximum of $\frac{1}{(1+(x- \frac{1}{2})^2 )^{\frac{3}{2}}}+\frac{1}{(1+(x+ \frac{1}{2})^2 )^{\frac{3}{2}}}$

Question

How to find this function's maximum?

$$\frac{1}{(1+(x- \frac{1}{2})^2 )^{\frac{3}{2}}}+\frac{1}{(1+(x+ \frac{1}{2})^2 )^{\frac{3}{2}}}$$

I think it has a maximum value at x=0. A search on Wolfram Alpha revealed that this is correct. However, the result in wolframalpha was to solve the 12th-order equation by differentiating and then performing general differentiation, I wonder if there is an easier way.

Answer 1

If you set \begin{align} f(x) = \frac{1}{\left(1+\left(x-\tfrac{1}{2}\right)^2\right)^{3/2}} +\frac{1}{\left(1+\left(x+\tfrac{1}{2}\right)^2\right)^{3/2}}, \end{align} there are a few quick observations. $f$ is an even function defined and continuously differentiable for all $x$, it clearly decreases for $x > \tfrac{1}{2},$ has a local maximum or minimum when $x=0$ and has limit zero when $x \to \pm \infty$. Thus the question reduces to whether there is a (greater) local maximum in the interval $\left(0,\tfrac{1}{2}\right]$.

We can calculate the first derivative, \begin{align} f'(x) &= -\frac{3}{2}\frac{ (2x-1)\left(1+\left(x+\tfrac{1}{2}\right)^2\right)^{5/2} +(2x+1)\left(1+\left(x-\tfrac{1}{2}\right)^2\right)^{5/2} } {\left(1+\left(x+\tfrac{1}{2}\right)^2\right)^{5/2}\left(1+\left(x-\tfrac{1}{2}\right)^2\right)^{5/2}} \\ &= -\frac{3}{2}\frac{\left(x^2 + \frac{5}{4}\right)^{5/2}} {\left(1+\left(x+\tfrac{1}{2}\right)^2\right)^{5/2} \left(1+\left(x-\tfrac{1}{2}\right)^2\right)^{5/2}} \\ &\hspace{1cm} \cdot \left( (2x-1)\left(1+\frac{x}{x^2+5/4}\right)^{5/2} +(2x+1)\left(1-\frac{x}{x^2+5/4}\right)^{5/2} \right)\tag{1} \end{align} To determine its sign, because the other $x$-terms are positive, we need only concentrate on the rightmost term, \begin{align} (2x-1)\left(1+\frac{x}{x^2+5/4}\right)^{5/2} +(2x+1)\left(1-\frac{x}{x^2+5/4}\right)^{5/2}. \end{align} For $x \in [0,1/2]$ the first term is negative and the second positive, and because $x/(x^2+5/4) < \tfrac{4}{5}x $, the rightmost term is bounded below by \begin{align} g(x) = (2x-1)\left(1+\tfrac{4}{5}x\right)^{5/2} +(2x+1)\left(1-\tfrac{4}{5}x\right)^{5/2}. \end{align} Now $g(0) =0$ and, over the interval $(0,1/2]$, \begin{align} g'(x) &= 2\left(1+\tfrac{4}{5}x\right)^{5/2}+(2x-1)\cdot \tfrac{5}{2}\tfrac{4}{5}(1+\tfrac{4}{5}x)^{3/2}\\ &\hspace{1cm} + 2\left(1-\tfrac{4}{5}x\right)^{5/2}-(2x+1)\cdot \tfrac{5}{2}\tfrac{4}{5}(1-\tfrac{4}{5}x)^{3/2}\\ &=\tfrac{28}{5}x\left(1+\tfrac{4}{5}x\right)^{3/2}-\tfrac{28}{5}x\left(1-\tfrac{4}{5}x\right)^{3/2}\\ &= \tfrac{28}{5}x \left( \left(1+\tfrac{4}{5}x\right)^{3/2}-\left(1-\tfrac{4}{5}x\right)^{3/2}\right)\\ &> 0 \end{align} Thus $g$ is strictly increasing so that $g(x) > 0$ in this interval. It follows $f'(x) < 0$ so that $f$ is decreasing in $(0,1/2]$. We observed earlier that $f(x)$ continues to decrease when $x \geq 1/2$. Thus $f(0)$ is the global maximum.

Answer 2

Hint :

Let :

$$f(x)=\frac{1}{(1+(x- \frac{1}{2})^2 )^{\frac{3}{2}}}+\frac{1}{(1+(x+ \frac{1}{2})^2 )^{\frac{3}{2}}}$$

We find the antiderivative of $f(x)$ wich is :

$$F(x)=g(x)-g(-x)$$

Where :

$$\frac{2\left(x+1\right)}{\sqrt{4x^{2}+2x+5}}=g(x)$$

Now we want to show that :

$$h(x)=F(x)-\frac{16}{5\sqrt{5}}x$$

Is decreasing on $x\in[0,\infty)$

The trick is to split in two the problem introducing a new function such that :

$$u(x)=g(x)-\frac{8}{5\sqrt{5}}\left(x+\frac{a}{x^{2}+1}\right)$$

And :

$$v(x)=-g(-x)-\frac{8}{5\sqrt{5}}\left(x-\frac{a}{x^{2}+1}\right)$$

Where $a\simeq 0.674$ and such that the two functions $u(x)$ and $v(x)$ are decreasing and admits a single extrema at $x=0$

The result follows .