Third-degree cosine inequality for obtuse triangle Suppose $\triangle ABC$ is an obtuse triangle with side lengths $a=BC, b=CA, c=AB$. I want to show that $$a^3\cos A+b^3\cos B+c^3\cos C<abc.$$
My idea is to use the cosine rule. I have $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, etc. Plugging into the inequality I get
$$a^4b^2+a^4c^2-a^6+b^4a^2+b^4c^2-b^6+c^4a^2+c^4b^2-c^6<2a^2b^2c^2.$$
How can I show this?
 A: Using $\sin2x,\cos2x$ formula
$$a^3\cos A=(2R\sin A)^3\cos A= 2R^3(2\sin^2A)(2\sin A\cos A)$$
$$=2R^3(1-\cos2A)\sin2A=R^3(2\sin2A-2\sin2A\cos2A)=R^3(2\sin2A-\sin4A)$$
Using this, $\sum \sin2A=4\prod \sin A$
Now, $\sin4A+\sin4B+\sin4C=2\sin(2A+2B)\cos(2A-2B)+2\sin2C\cos2C$
Now, $\cos2C=\cos\{2\pi-2(A+B)\}=\cos2(A+B)$ and $\sin2(A+B)=\sin(2\pi-2C0=-\sin2C$
$\implies\sin4A+\sin4B+\sin4C=-2\sin2C\cos(2A-2B)+2\sin2C\cos2(A+B)$
$=-\sin2C\{\cos(2A-2B)-\cos(2A+2B)\}=-\sin2C\cdot2\sin2A\sin2B$
$=-2(2\sin C\cos C)(2\sin A\cos A)(2\sin B\cos B)$
For  an obtuse triangle, only one angle is between $(\frac\pi2,\pi)$ so exactly one of the cosine ratio $<0$ and all the sine ratios are $>0$
$\implies\sin4A+\sin4B+\sin4C>0$
$\implies\sum a^3\cos A <2R^3(\sum\sin2A)=2R^3(4\sin A\sin B\sin C)=\prod(2R\sin A)$
A: here is another way follow op's idea:
for easy, $x=a^2,y=b^2,z=c^2,$ WOLG, let $C$ is obtuse triangle, then $ x+y<z$, we want to prove :
$x^2y+x^2z-x^3+y^2x+y^2z-y^3+z^2x+z^2y-z^3<2xyz$
that is to prove: when $z>x+y$
$f(z)=-(x-y)^2(x+y)+(x-y)^2z+(x+y)z^2-z^3 <0$
now we prove $f(z)$ is mono decreasing function:
$f'(z)=(x-y)^2+2(x+y)z-3z^2$
$f''(z)=2(x+y)-6z<0 \implies f'(z)<f'(x+y)=-4xy<0 \implies f(z) <f(x+y)=0$  
