Prove or disprove the implication:

I will follow the method suggested by Blue in his comment to show that the implication does not hold.

The sides $a$, $b$, $c$ are proportional to $\sin A$, $\sin B$ and $\sin C$, respectively, by the sine law, so the equation is equivalent to $S = 0$, where $$ S = \sin^2 A \tan(B-C) + \sin^2 B \tan(C-A) + \sin^2 C \tan(A - B).$$ Letting $x = \tan A$, $y = \tan B$, and $z = \tan C$, we can rewrite $S$ in terms of $x$, $y$, $z$; for example, we have $\sin^2 A = x^2/(1 + x^2)$ and $\tan(B - C) = (y-z)/(1 + yz)$. (We will ignore for now the special case where $ABC$ is a right triangle, and one of $x$, $y$, $z$, is correspondingly infinite.) Thus $$ S = \frac{x^2}{1 + x^2} \frac{y-z}{1 + yz} + \frac{y^2}{1 + y^2} \frac{z-x}{1 + zx} + \frac{z^2}{1 + z^2} \frac{x-y}{1 + xy.}$$ Now we note that $S = 0$ when any two of $x, y, z$ are equal, so we expect the numerator of $S$ to be divisible by $(y-z)(z-x)(x-y)$. This is indeed the case (I used a computer for this part, although it is possible to do long division first by $y-z$, then by $z-x$ and finally by $x-y$): $$S = \frac{(y-z)(z-x)(x-y)(2x^2 y^2 z^2 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 1)}{(1 + yz)(1 + zx)(1 + xy)(1 + x^2)(1 + y^2)(1 + z^2)}.$$ Now that we have recovered the factors $y-z$, $z-x$ and $x - y$, we can split off factors equal to the tangents that originally appeared in $S$, so that $$S = \frac{y-z}{1 + yz} \frac{z-x}{1 + zx} \frac{x-y}{1 + xy} T = \tan(B-C)\tan(C-A)\tan(A-B) T,$$ where $$ T = \frac{2x^2 y^2 z^2 + x^2 y^2 + y^2 z^2 + z^2 x^2 - 1}{(1 + x^2)(1 + y^2)(1 + z^2)}.$$ To simplify $T$, we write $X = 1 + x^2$, $Y = 1 + y^2$, $Z = 1 + z^2$, so that $X = 1 + \tan^2 A = \sec^2 A$, and similarly, $Y = \sec^2 B$, $Z = \sec^2 C$. Then by replacing $x^2$, $y^2$ and $z^2$ with $X-1$, $Y-1$, $Z-1$, respectively, where they appear in $T$, we obtain $$ \begin{align} T &= \frac{2(X-1)(Y-1)(Z-1) + (X-1)(Y-1) + (Y-1)(Z-1) + (Z-1)(X-1) - 1}{XYZ} \\ &= \frac{2XYZ - XY - YZ - ZX}{XYZ} \\ &= 2 - \frac{1}{X} - \frac{1}{Y} - \frac{1}{Z} \\ &= 2 - \cos^2 A - \cos^2 B - \cos^2 C. \end{align} $$ We therefore have the following identity: $$S = \tan(B-C)\tan(C-A)\tan(A-B) (2 - \cos^2 A - \cos^2 B - \cos^2 C),$$ which corresponds more or less to Blue's factorization. Interestingly, we never used the fact that $A$, $B$ and $C$ were the angles of a triangle. The identity extends to the case where one of the angles is a right angle by continuity.

Now the only question becomes whether it is possible to find a scalene triangle for which $$\cos^2 A + \cos^2 B + \cos^2 C = 2.$$ If we assume that $A < B < C$, then this amounts to showing that the function $$f(A,B) = \cos^2 A + \cos^2 B + \cos^2 (A + B)$$ takes the value $2$ somewhere in the interior of the triangle $\Delta$ bounded by the lines $A = 0$, $B = A$ and $A + 2B = \pi$. At the vertex $(0,0)$ of $\Delta$, we find $f(0,0) = 3$, so by continuity, at points in the interior of $\Delta$ near $(0,0)$, we must have $f(A,B) \approx 3$. Likewise, at the vertex corresponding to an equilateral triangle we have $f(\pi/3, \pi/3) = 3/4$, so at some points in the interior of $\Delta$, we must have $f(A, B) \approx 3/4$. Since the interior of $\Delta$ is a connected set, by the intermediate value theorem, we must have $f(A, B) = 2$ for some choice of $(A, B)$ in the interior of $\Delta$, which therefore corresponds to a scalene triangle.

Question: Was it predictable somehow, before the factoring step, that the numerator of $T$ would be expressible in terms of $x^2$, $y^2$ and $z^2$? That is what allowed the method to work as smoothly as it did.

The claim holds if we replace the function $\tan(B-C)$ (and cyclic permutations) with $\tan^*(B-C)=\frac{\sin(B-C)}{\cos(B+C)}$: Since: $$\tan^*(B-C)=\frac{\sin B \cos C - \sin C \cos B}{\cos B \cos C - \sin B \sin C}$$ and $2ab\cos(C) = a^2+b^2-c^2, 2R\sin A=a, 2ab\sin C=4\Delta$, we have: $$2R\tan^*(B-C)= \frac{2abc(a^2+b^2-c^2-a^2-c^2+b^2)}{(a^2+c^2-b^2)(a^2+b^2-c^2)-16\Delta^2},$$ but Heron's formula gives $16\Delta^2 = 4b^2c^2-(b^2+c^2-a^2)^2$, from which $$2R\tan^*(B-C) = \frac{4abc(b^2-c^2)}{a^4-(b^2-c^2)^2-4b^2c^2+(b^2+c^2-a^2)^2}$$ follows. This gives: $$2R\tan^*(B-C) = \frac{2abc(b^2-c^2)}{a^2(a^2-b^2-c^2)}=\frac{c^2-b^2}{a \cos A}$$ and: $$2Ra^2\tan^*(B-C) = \frac{a(c^2-b^2)}{\cos A} = 2R(c^2-b^2)\tan(A),$$ $$ a^2 \tan^*(B-C) = (c^2-b^2)\tan A = (c^2-b^2)\frac{4\Delta}{b^2+c^2-a^2}.\tag{1}$$ So we have, in terms of side lengths, $$ \sum_{cyc}(c^2-b^2)(a^2+c^2-b^2)(a^2+b^2-c^2) = 0, \tag{2} $$ or: $$ \sum_{cyc}(c^2-b^2)(a^4-b^4-c^4+2b^2c^2) = 0,$$ $$ \sum_{cyc}\left(b^6-c^6-a^4b^2+a^2b^4+3b^2c^4-3b^4c^2\right)=0, $$ $$ \sum_{cyc}\left(-a^4b^2+a^2b^4+3b^2c^4-3b^4c^2\right)=0, $$ $$ \sum_{cyc}\left(4b^2c^4-4b^4c^2\right)=0,\tag{3} $$

Now it is easy to notice that $a=\pm b,a=\pm c$ or $b=\pm c$ imply that the LHS is zero, so $(a^2-b^2)(b^2-c^2)(c^2-a^2)$ divides the LHS in $(3)$.

$$ \sum_{cyc}\left(4b^2c^4-4b^4c^2\right) = 4(a^2-b^2)(b^2-c^2)(c^2-a^2)\tag{4} $$ proves the (modified) claim.