<s>[INST] 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?
 [/INST] 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.)
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