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DrunkenOldFool
New member
- Feb 6, 2012
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$a,b,c$ are any three positive numbers such that $a+b+c=1$. Prove that
$$ab^2c^3 \leq \frac{1}{432}$$
$$ab^2c^3 \leq \frac{1}{432}$$
Problem involving arithmetic and geometric mean.
- Thread starter DrunkenOldFool
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sbhatnagar
Active member
- Jan 27, 2012
- 95
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.
$\displaystyle GM=\left( \frac{b}{2}\frac{b}{2}\frac{c}{3}\frac{c}{3}\frac{c}{3}\right)^{\frac{1}{6}}=\left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$
$AM \geq GM$
$\displaystyle \frac{1}{6} \geq \left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$
$\displaystyle \Rightarrow \frac{2^23^3}{6^6} \geq ab^2c^3$
$$a,\frac{b}{2},\frac{b}{2},\frac{c}{3},\frac{c}{3},\frac{c}{3}$$
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.
$\displaystyle GM=\left( \frac{b}{2}\frac{b}{2}\frac{c}{3}\frac{c}{3}\frac{c}{3}\right)^{\frac{1}{6}}=\left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$
$AM \geq GM$
$\displaystyle \frac{1}{6} \geq \left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$
$\displaystyle \Rightarrow \frac{2^23^3}{6^6} \geq ab^2c^3$