Can there be a triangle ABC if $\frac{\cos A}{1}=\frac{\cos B}{2}=\frac{\cos C}{3}$? Can there be a triangle ABC if  $$\frac{\cos A}{1}=\frac{\cos B}{2}=\frac{\cos C}{3}\;?$$ Equating the ratios to $k$ we get $\cos A=k$, $\cos B=2k$, $\cos C=3k$.
Then the identity $$\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1 \implies 12k^3+14k^2-1=0$$
$f(k)=12k^3+14k^2-1$ being monotonic for $k>0$ can have at most one real positive root. Further, $f(0)=-1, f(1/3)=1>0$, so there will be one real root in $(0,1/3)$. Hence all three cosines will be positive and less that 1, for a unque triangle to be possible.
What can be other ways to solve  this question?
 A: The function $f(k) = \cos^{-1}(k) + \cos^{-1}(2k) + \cos^{-1}(3k)$ is continuous on $k\in [0,1/3]$. Since $f(0) = 3\pi/2 > \pi$ and $f(1/3) \approx 2.07 < \pi$, the intermediate value theorem tells us that there is a $k \in (0,1/3)$ such that $f(k) = \pi$. This is the desired triangle.
A: \begin{align} 
\frac{\cos A}{1}&
=\frac{\cos B}{2}=\frac{\cos C}{3}
=\frac{\cos A+\cos B+\cos C}{6}
\tag{1}\label{1}
.
\end{align}
\begin{align} 
6\cos A&=\cos A+\cos B+\cos C
=\frac{r}{R}+1=
v+1
\tag{2}\label{2}
,
\end{align}
where $r$ and $R$ are inradius and circumradius
of the corresponding $\triangle ABC$,
$v=\tfrac{r}{R}\in(0,\tfrac12]$ for a valid triangle.
Hence,
\begin{align} 
\cos A&=\frac{v+1}6\in(0,\tfrac14)
\tag{3}\label{3}
,\\
\cos B&=2\cos A =\frac{v+1}3 \in(0,\tfrac12)
\tag{4}\label{4}
,\\
\cos C&=3\cos A=\frac{v+1}2 \in(0,\tfrac34)
\tag{5}\label{5}
.
\end{align}
From known identity
\begin{align} 
\cos A\cos B\cos C
&=\tfrac14(u^2-(v+2)^2)
\tag{6}\label{6}
,
\end{align}
where $u=\tfrac\rho{R}$,
and $\rho$ is semiperimeter of $\triangle ABC$,
we can express $u^2$ in terms of $v$:
\begin{align} 
\tfrac1{36}(v+1)^3
&=\tfrac14(u^2-(v+2)^2)
,\\
u^2&=
\tfrac19v^3+\tfrac43v^2+\tfrac{13}3 v+\tfrac{37}9
\tag{7}\label{7}
,
\end{align}
and with the help of
\begin{align} 
\cos A\cos B+\cos B\cos C+\cos C\cos A
&= \tfrac14(u^2+v^2)-1
\tag{8}\label{8}
\end{align}
we arrive at the cubic equation in $v$
\begin{align}
v^3+10v^2+17v-10&=0
,
\end{align}
which has only one positive solution
\begin{align}
v&=
\tfrac{14}3\cos\left(
\tfrac\pi3
-\tfrac13\arctan\left(\tfrac9{100}\sqrt{1329}\right)
\right)
-\tfrac{10}3
\approx 0.458757
.
\end{align}
Then
\begin{align}
u(v) &= \tfrac13\sqrt{v^3+12v^2+39v+37}
\approx 2.52792343
,\\
u_{\min}(v)&=
\sqrt{27-(5-v)^2-2\sqrt{(1-2v)^3}}
\approx 2.5158959
,\\
u_{\max}(v)&=
\sqrt{27-(5-v)^2+2\sqrt{(1-2v)^3}}
\approx 2.53465839
,
\end{align}
so indeed we have a unique valid triangle with given properties.
A: Well, not as sophisticated.
But for any angle $\frac \pi 2 > A > \frac \pi 3$ we can have $\cos A$ be any value from $0 < \cos A < \frac 1 2$ we can have $B_A= \cos^{-1} (2\cos A)$ so $\cos B_A = 2\cos A$ and $\frac \pi 2 > B_A > 0$ and $0 < \cos B < 1$.
These can be two angles of a triangle with the third angle being $C_A= \pi - A - B_A=\pi -A - \cos^{-1}(2\cos A)$.
We need $\cos C_A = 3\cos A$ or $\cos C_A - 3\cos A = 0$.  Is that possible?
Well,  If $A = \frac \pi 2$ then $\cos A =0$ and $B_A = \cos^{-1} 0 = \frac \pi 2$ and $C_A = 0$ and $\cos C_A = 1$ and $\cos C_A - 3\cos A = 1$.
And If $A=\frac \pi 3$ then $\cos A = \frac 12$ and $B_ = \cos^{-1} 1 = 0$ and $C_A = \frac {2\pi}3$ and $\cos C_A = -\frac 12 $ and $\cos C_1 - 3\cos A = -2$.
But $\cos C_A -3\cos A$ is continuous, so there must be a value of $A$ between $\frac \pi 2$ and $\frac \pi 3$ where $\cos C_A - 3\cos A = 0$.
And for that $A$ we will have $B_A = \cos^{-1}(2\cos A)$ and $\cos B_A = 2\cos A$ and $\cos C_A = 3\cos A$.
Ta-da....
(You never said you wanted to find it; just that you wanted to know one exists.  Well, one must exist by continuity and intermediate value theorem but this argument gives utterly no method of finding it.)
