Solve $\arctan(ax)+\arctan(bx)+\arctan(cx) = \pi$ as $x=\frac{\sqrt{a+b+c}}{\sqrt{abc}}$

Question

I know that is made applying the cosine and with a inverse trigonometric relationship, but I'm stuck after that.

Answer 1

HINT:

$$\tan\{\arctan(ax)+\arctan(bx)+\arctan(cx)\}=\dfrac{(a+b+c)x-abcx^3}{1-x^2(ab+bc+ca)}$$

For $x=\dfrac{\sqrt{a+b+c}}{\sqrt{abc}},$

$$(a+b+c)x-abcx^3=(a+b+c)\cdot\dfrac{\sqrt{a+b+c}}{\sqrt{abc}}-abc\cdot\dfrac{(a+b+c)^{3/2}}{(abc)^{3/2}}=0$$

$\implies \arctan(ax)+\arctan(bx)+\arctan(cx)=n\pi$ where $n$ is any integer

As $\arctan(y)<\dfrac\pi2$ for real finite $y$ we need $ax,bx,cx>0$

otherwise $\arctan(ax)+\arctan(bx)+\arctan(cx)<\pi$

If $a,b,c>0;$ we need $x>0$

As $abc>0,$ even number of terms in $\in\{a,b,c\}$ must be negative.

But we need $a,b,c$ of same sign as $x\implies a,b,c$ must be $>0$

Answer 2

let $\tan \alpha = ax$, $\tan \beta = bx$, $\tan \theta = cx$.

$$ \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{a x + b x}{1 - ab x^2} $$

Substituting in $x= \frac{\sqrt{a+b+c}}{\sqrt{abc}}$ and working through leads to

$$ \tan (\alpha + \beta) = \frac{- c \sqrt{a+b+c} }{\sqrt{abc}} = - c x = -\tan{\theta} $$

which provides the result that $\alpha + \beta + \theta = \pi$ if we assume that $a,b,c$ (and hence $x$) are positive and finite, i.e. that $0 < \alpha, \beta, \theta < \pi/2$.

Answer 3

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I'll assume $\ds{a, b, c \in \mathbb{R}_{\ >\ 0}}$

\begin{align} x & = {\root{a + b + c} \over \root{abc}} \implies abcx^{2} = a + b + c \implies abcx^{3} - cx = ax + bx \\[5mm] \implies cx &= {ax + b x \over abx^{2} - 1} = {1/\pars{bx} + 1/\pars{ax} \over 1 - \bracks{1/\pars{bx}}\bracks{1/\pars{ax}}} \\[5mm] \implies \arctan\pars{cx} & = \arctan\pars{1 \over bx} + \arctan\pars{1 \over ax} \\[5mm] \implies \arctan\pars{cx} & = \bracks{{\pi \over 2} - \arctan\pars{bx}} + \bracks{{\pi \over 2} - \arctan\pars{ax}} \\[5mm] \implies &\bbx{\ds{\arctan\pars{ax} + \arctan\pars{bx} + \arctan\pars{cx} = \pi}} \end{align}