Proving Inequality: $a^{2a}b^{2b}c^{2c} > a^{b+c}b^{c+a}c^{a+b}$

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The inequality \( a^{2a}b^{2b}c^{2c} > a^{b+c}b^{c+a}c^{a+b} \) is proven under the condition that \( a > b > c > 0 \). The discussion confirms the validity of this inequality through mathematical reasoning. Participants agree on the correctness of the proof, emphasizing the importance of the ordering of the variables.

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given:$a>b>c>0$

prove:$a^{2a}b^{2b}c^{2c}>a^{b+c}b^{c+a}c^{a+b}$
 
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Albert said:
given:$a>b>c>0$

prove:$a^{2a}b^{2b}c^{2c}>a^{b+c}b^{c+a}c^{a+b}$

Dividing both sides of the inequality by the expression on the right gives an equivalent inequality

$\displaystyle \left(\frac{a}{b}\right)^{a-b} \left(\frac{b}{c}\right)^{b-c} \left(\frac{a}{c}\right)^{a-c} > 1$,

which holds since the bases $\frac{a}{b}, \frac{b}{c}, \frac{a}{c}$ are all greater than 1 and the exponents $a-b, b-c, a-c$ are all positive.
 
Euge said:
Dividing both sides of the inequality by the expression on the right gives an equivalent inequality

$\displaystyle \left(\frac{a}{b}\right)^{a-b} \left(\frac{b}{c}\right)^{b-c} \left(\frac{a}{c}\right)^{a-c} > 1$,

which holds since the bases $\frac{a}{b}, \frac{b}{c}, \frac{a}{c}$ are all greater than 1 and the exponents $a-b, b-c, a-c$ are all positive.
very good ,you got it !
 

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