<s>[INST] For $\triangle ABC$, prove $( \sin A + \sin B )( \sin B +\sin C )( \sin C + \sin A) > \sin A \sin B \sin C$ 
In $\triangle$ ABC, prove that
  $$( \sin A + \sin B )( \sin B + \sin C )( \sin C + \sin A) > \sin A \sin B \sin C$$

I have tried the formula A.M.- G.M. relation with $\sin A$, $\sin B$, and $\sin C$, but it isn't really helping. So, what is the correct proof?
 [/INST] Let the sides of the triangle be $a,b,c.$ Let $R$ be the circumradius of the triangle  We have $$0< 2 R=(\sin A)/a=(\sin B)/b=(\sin C)/c.$$ So the inequality is equivalent to $$(a+b)(b+c)(c+a)>a b c.$$ 
In a triangle we have $a+b>c>0$ and $b+c>a>0$ and $c+a>b>0 .$ So $(a+b)(b+c)(c+a)>c a b.$
Alternate proof: In a triangle the sines of the angles are all positive. If you expand your original LHS you have 8 positive terms and two of those terms are each equal to the RHS!
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