MHB Proof of $a^{2a}\times b^{2b}\times c^{2c}>a^{b+c}\times b^{c+a}\times c^{a+b}$

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    Inequality Proof
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The discussion centers on proving the inequality \( a^{2a} \times b^{2b} \times c^{2c} > a^{b+c} \times b^{c+a} \times c^{a+b} \) under the condition that \( a > b > c > 0 \). Participants explore various mathematical approaches and techniques to establish this inequality, emphasizing the significance of the relationship between the variables. Several hints and strategies are proposed, including the application of logarithmic properties and inequalities. The conversation highlights the complexity of the proof and the need for a rigorous mathematical foundation. Ultimately, the goal is to validate the inequality through logical reasoning and mathematical principles.
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given :
$a>b>c>0$
prove :
$a^{2a}\times b^{2b}\times c^{2c}>a^{b+c}\times b^{c+a}\times c^{a+b}$
 
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Albert said:
given :
$a>b>c>0$
prove :
$a^{2a}\times b^{2b}\times c^{2c}>a^{b+c}\times b^{c+a}\times c^{a+b}$
hint:
let:
$A=a^{2a}\times b^{2b}\times c^{2c}$
$B=a^{b+c}\times b^{c+a}\times c^{a+b}$
prove:$\dfrac {A}{B}>1$
 
Albert said:
hint:
let:
$A=a^{2a}\times b^{2b}\times c^{2c}$
$B=a^{b+c}\times b^{c+a}\times c^{a+b}$
prove:$\dfrac {A}{B}>1$
for $a>b>c>0$ we have :
$\dfrac {A}{B}=a^{(a-b)}a^{(a-c)}b^{(b-a)}b^{(b-c)}c^{(c-a)}c^{(c-b)}$
$=(\dfrac{a}{b})^{a-b}(\dfrac{b}{c})^{b-c}(\dfrac{a}{c})^{a-c}>1^0\times1^0\times 1^0=1$
 
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