MHB Problem involving arithmetic and geometric mean.

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For three positive numbers \(a, b, c\) such that \(a + b + c = 1\), it is proven that \(ab^2c^3 \leq \frac{1}{432}\). The proof utilizes the arithmetic mean (AM) and geometric mean (GM) inequality by considering the numbers \(a, \frac{b}{2}, \frac{b}{2}, \frac{c}{3}, \frac{c}{3}, \frac{c}{3}\). The AM is calculated to be \(\frac{1}{6}\), while the GM is expressed in terms of \(ab^2c^3\). The inequality \(AM \geq GM\) leads to the conclusion that \(\frac{2^2 3^3}{6^6} \geq ab^2c^3\), confirming the original statement. This establishes a clear relationship between the means and the product of the variables.
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$a,b,c$ are any three positive numbers such that $a+b+c=1$. Prove that

$$ab^2c^3 \leq \frac{1}{432}$$
 
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Consider the 6 numbers

$$a,\frac{b}{2},\frac{b}{2},\frac{c}{3},\frac{c}{3},\frac{c}{3}$$

The arithmetic mean of these numbers is

$\displaystyle AM = \dfrac{a+\frac{b}{2}+\frac{b}{2}+\frac{c}{3}+\frac{c}{3}+\frac{c}{3}}{6}$

$=\frac{1}{6}$

Similarly, you can calculate the Geometric Mean.

$\displaystyle GM=\left( \frac{b}{2}\frac{b}{2}\frac{c}{3}\frac{c}{3}\frac{c}{3}\right)^{\frac{1}{6}}=\left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$

$AM \geq GM$
$\displaystyle \frac{1}{6} \geq \left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$

$\displaystyle \Rightarrow \frac{2^23^3}{6^6} \geq ab^2c^3$
 

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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