MHB 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 to be proven under the condition that \( a > b > c > 0 \). The discussion confirms that the inequality holds true given the specified conditions. Participants express satisfaction with the proof provided. The focus remains on the mathematical validation of the inequality rather than alternative perspectives. The conclusion emphasizes the correctness of the inequality under the stated assumptions.
Albert1
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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 !
 

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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