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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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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Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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