Inequality

(a+b)(b+c)(c+a) ≥ 8abc

For any a, b and c 0 we have

(a + b)(b + c)(c + a) ≥ 8abc

 

Proof. We have

inequality

inequality

We add 4ab to this

inequality

And now we have the same thing here

 

a, b and c are non-negative numbers.

So, we can multiply these 3 inequalities:

inequality

inequality (a+b)(b+c)(c+a)>8abc

 

Keywords: Algebraic Inequalities, algebra, inequality

 

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