Proving the Inequality ( frac{ab}{c} cdot frac{bc}{a} cdot frac{ca}{b} geq abc )
The statement ( frac{a b}{c} cdot frac{b c}{a} cdot frac{c a}{b} geq abc ) is a mathematical inequality that can be proven by several methods. Here, we will discuss a detailed proof using the Arithmetic Mean-Geometric Mean (AM-GM) Inequality as a key tool.
Proof Using AM-GM Inequality
Let's start by defining the following variables for simplicity:
C ( frac{ab}{c} ) A ( frac{bc}{a} ) B ( frac{ca}{b} )Then the left side of the inequality becomes ( ABC ).
Step 1: Simplify the Right Side
The right side can be simplified as follows:
Calculate ( A cdot B ): ( A cdot B left( frac{bc}{a} right) left( frac{ca}{b} right) frac{bc cdot ca}{ab} c^2 ) Therefore, ( c sqrt{AB} ) Substitute ( c sqrt{AB} ) into the right side of the inequality: ( frac{a b}{c} cdot frac{b c}{a} cdot frac{c a}{b} AB cdot frac{a b}{sqrt{AB}} cdot frac{b c}{a} cdot frac{c a}{b} AB cdot frac{a b}{sqrt{AB}} cdot c ) Since ( c sqrt{AB} ), this becomes: ( AB cdot frac{a b}{sqrt{AB}} cdot sqrt{AB} AB cdot ab (AB cdot ab) / 1 (a b) (b c) (c a)/abc abc )Thus, we have shown that ( frac{ab}{c} cdot frac{bc}{a} cdot frac{ca}{b} geq abc ).
Equality Condition
Equality holds if and only if ( A B C ), which implies that ( a b c ).
Alternative Proof Using AM-GM Inequality
We can also prove this inequality using the AM-GM Inequality, which states that for any non-negative real numbers ( x, y, z ), the following holds:
( x^2y^2 geq 2xy )
Applying the AM-GM Inequality, we get:
( x^2y^2 geq 2xy ) ( y^2z^2 geq 2yz ) ( z^2x^2 geq 2xz )Adding these inequalities, we obtain:
( x^2y^2 y^2z^2 z^2x^2 geq 2xy 2yz 2xz )
Dividing by ( xyz ) on both sides, we get:
( frac{x^2y^2}{xyz} frac{y^2z^2}{xyz} frac{z^2x^2}{xyz} geq frac{2xy}{xyz} frac{2yz}{xyz} frac{2xz}{xyz} )
Which simplifies to:
( frac{xy}{z} frac{yz}{x} frac{zx}{y} geq 2 left( frac{y}{z} frac{z}{y} frac{x}{z} right) )
Dividing both sides by ( xyz ), we get the desired result:
( frac{ab}{c} cdot frac{bc}{a} cdot frac{ca}{b} geq abc )
Conclusion
In conclusion, the inequality ( frac{ab}{c} cdot frac{bc}{a} cdot frac{ca}{b} geq abc ) holds true, and equality is achieved when ( a b c ).