Find the minimum positive value that $(f(x)-g(x)+3)$ may attain.

Question

Let $f:\mathbb R\to \mathbb R$ be two non-constant differentiable functions. If $f'(x)-g'(x)=g'(x)(f(x))^2-f'(x)(g(x))^2$ for all $x\in\mathbb R$ and $f(1)=1, g(1)=\frac13$ and if there is no $x$ for which $f(x)=-2$ or $g(x)=2$ then find the minimum positive value that $(f(x)-g(x)+3)$ may attain.

My Attempt:

$f'(x)(1+(g(x)^2)=g'(x)(1+(f(x))^2)$

$\frac{f'(x)}{1+(f(x))^2}=\frac{g'(x)}{1+(g(x))^2}$

$\big[\arctan(f(x))\big]_1^x=\big[\arctan(g(x))\big]_1^x$

$\arctan(f(x))-\arctan1=\arctan(g(x))-\arctan\frac13$

$\arctan(f(x))-\arctan(g(x))=\arctan1-\arctan\frac13=\arctan\frac12$

Not sure what to do next.

Perhaps the answer is $2√5-1$.

Answer 1

You've made great progress. Using the Arctangent addition formula, along with your result of $\arctan(f(x)) - \arctan(g(x)) = \arctan\left(\frac{1}{2}\right)$, we get that

$$\arctan\left(\frac{f(x) - g(x)}{1 + f(x)g(x)}\right) = \arctan\left(\frac{1}{2}\right)$$

The arguments are congruent modulo $\pi$ so, for some fixed integer $n$, we have

$$\frac{f(x) - g(x)}{1 + f(x)g(x)} = n\pi + \frac{1}{2}$$

From $f(1) = 1$ and $g(1) = \frac{1}{3}$, we get $\frac{f(x) - g(x)}{1 + f(x)g(x)} = \frac{1}{2}$, so $n = 0$. Thus,

$$\begin{equation}\begin{aligned} \frac{f(x)-g(x)}{1+f(x)g(x)} &= \frac{1}{2} \\ 2f(x) - 2g(x) & = 1 + f(x)g(x) \\ -f(x)g(x) + 2f(x) - 2g(x) - 1 &= 0 \\ -f(x)g(x) + 2f(x) - 2g(x) + 4 - 4 - 1 & = 0 \\ (f(x) + 2)(-g(x) + 2) & = 5 \end{aligned}\end{equation}$$

Note that, with $f(1) = 1$ and $g(1) = \frac{1}{3}$, both of the LHS factors are positive. Also, they remain positive since $f(x) \neq -2$ and $g(x) \neq 2$ (this is likely why that restriction is given since, otherwise, the LHS becomes $0$). Thus, by the AM-GM inequality, we get

$$\frac{(f(x) + 2) + (-g(x) + 2)}{2} \ge \sqrt{5} \;\;\to\;\; f(x) - g(x) + 3 \ge 2\sqrt{5} - 1$$

This shows the minimum positive value that may be obtained is $2\sqrt{5} - 1$, as you suggested, with this occurring when $f(x) + 2 = -g(x) + 2 = \sqrt{5}$.

Answer 2

In what follows, $f(x)$ and $g(x)$ will be written as $f$ and $g$ respectively.

Note that

$$\tan^{-1} f = \tan^{-1} g + \tan^{-1} \left( \frac{1}{2} \right) $$ $$\implies f = \frac{g+ \frac{1}{2}}{1-\frac{g}{2}} $$ $$\implies f-g+3 = \frac{2g+1}{2-g}-g+3 $$ $$\implies f-g+3 = \frac{g^2-3g+7}{2-g}$$

Let $k=f-g+3$, then $k = \frac{g^2-3g+7}{2-g}$.

Therefore $k$ is a rational function of $g$.

The graph of $k = \frac{g^2-3g+7}{2-g}$ is plotted as below and the minumum can be found by standard differentiation method.

Alternatively, we may work as follows:

$$k = \frac{g^2-3g+7}{2-g}$$ $$\implies g^2-3g+7=2k-kg$$ $$\implies g^2+(k-3)g+(7-2k)=0$$

Since $g$ is always real, the discriminant of the above quadratic equation is non-negative.

$$(k-3)^2-4(7-2k) \geq 0$$ $$k^2+2k-19 \geq 0$$ $$(k+1)^2 \geq 20$$ $$k+1 \leq -2 \sqrt 5 \; \; \mathrm{or} \; \; k+1 \geq 2 \sqrt 5$$ $$k \leq -2\sqrt 5 -1 \; \; \mathrm{or} \; \; k \geq 2 \sqrt 5 - 1$$

Hence the minimum value of $k=f-g+3$ is $2 \sqrt 5 - 1$.