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?
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As long as this triangle is Euclidean, each of the sin values is positive (or maybe 0 but that case is easy). When you multiply these factors out, you will get $\sin A \sin B \sin C+$..(other positive stuff) which is definitely greater than $\sin A \sin B \sin C$
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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!
Why?
As $\sin A,\sin B,\sin C>0,$
$$\dfrac{\sin A+\sin B}2\ge\sqrt{\sin A\sin B}$$
$$\implies(\sin A+\sin B)(\sin B+\sin C)(\sin C+\sin A)\ge8\sin A\sin B\sin C$$ which is definitely $>\sin A\sin B\sin C$