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$$\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C)) $$
$$\implies \frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$
Given if $A>B>C$ and $A+B+C=180^\circ$
Is that implication correct if not then please correct it otherwise try to solve the first inequality.
This is not the original problem, but this problem arose when I was solving another trigonometric equation.
Now if anybody prove or disprove the above inequality then that will also be the solution of my problem.
Yes it is
B is always smaller than $ 90 ^\circ (A > B $ and $ 180^\circ/2 = 90^\circ )$
B is always greater than $ 0^\circ (B > 0 ) $
B-C is always greater than $ 0^\circ (B > C) $
B-C is smaller than $ 90 $ ( 1. and $C > 0$ )
So B-C is always between $ 0^\circ $ and $90^\circ$ ( $ 90^\circ > (B-C) > 0^\circ $ )
Therefore $ \tan(B-C) > 0 $
And you may divide both sides in inequalities by the same positive value.
The inequality is equivalent to: $$ \sin^2 A \sin(B-C)\cos(A-C) \geq \sin^2 B \sin(A-C)\cos(B-C),$$ or (using twice Briggs' formulas and $\cos(A+B)=-\cos(C)$): $$ \sin(3B-A)+\sin(2B+C)+\sin(2A-C)+\sin(2A-3C)\leq \sin(3A-B)+\sin(2A+C)+\sin(2B-C)+\sin(2B-3C)\tag{1},$$ that is equivalent to: $$ \sin(2A-2B)\cos(C)\geq\sin(A-B)\cos(4C)\tag{2} $$ or (using the sine duplication formula): $$2\cos(A-B)\cos C\geq \cos(4C).\tag{3}$$ Since $A>B>C>0$ and $A+B+C=\pi$ imply $A-B\leq\pi-3C$, we have: $$2\cos(A-B)\cos C\geq -2\cos C\cos(3C),$$ and equality holds when $A=\pi-2C,B=C$. In this case, $(3)$ holds only if $$2\cos C\cos(3C)+\cos(4C)\leq 0,$$ i.e. only if $C$ is big enough, for istance $$C\geq\operatorname{arccos}\left(\frac{1}{4}\sqrt{7+\sqrt{33}}\right)=0.4679647\ldots>\frac{7\pi}{47}.$$
The original inequality is $\sin^{2}A(\tan(B-C))>\sin^{2}B(\tan(A-C)) $
And we have to prove this $\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$. Now we have to interchange only the positions of $tan(B-C)$ and $sin^{2}B$. $sin^{2}B$ is always positive since it is a square. We need to care only about $tan(B-C)$.
If $tan(B-C)<0$ then $(B-C)>90^\circ \implies B>90^\circ+C$.
$C$ is a non-zero angle. Therefore $B>90^\circ$ and $A>B>90^\circ$.
Consider the fact that $A+B+C=180^\circ$. This means that the angles A, B , C form the angles of a triangle. But we know that there cannot exist two obtuse angles in a triangle. Therefore our supposition is wrong and hence $B-C<90^\circ$ which leads to $tan(B-C)>0$.
Now we can interchange $tan(B-C)$ and $sin^{2}B$ without changing the inequality sign. Hence we have the given inequality $$\frac{\sin^2 A}{\sin^2 B} > \frac{\tan(A-C)}{\tan(B-C)}$$
It's correct if $\tan(B-C)>0$, since then you're multiplying both sides by a positive number (the reciprocal of that tangent), unless $\sin^2 B=0$, in which case you've got other problems.
