Nov 17, 2012 Thread starter #1 D DrunkenOldFool New member Feb 6, 2012 20 $a,b,c$ are any three positive numbers such that $a+b+c=1$. Prove that $$ab^2c^3 \leq \frac{1}{432}$$
$a,b,c$ are any three positive numbers such that $a+b+c=1$. Prove that $$ab^2c^3 \leq \frac{1}{432}$$
Nov 17, 2012 #2 S 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. $\displaystyle GM=\left( \frac{b}{2}\frac{b}{2}\frac{c}{3}\frac{c}{3}\frac{c}{3}\right)^{\frac{1}{6}}=\left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$ $AM \geq GM$ $\displaystyle \frac{1}{6} \geq \left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$ $\displaystyle \Rightarrow \frac{2^23^3}{6^6} \geq ab^2c^3$
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. $\displaystyle GM=\left( \frac{b}{2}\frac{b}{2}\frac{c}{3}\frac{c}{3}\frac{c}{3}\right)^{\frac{1}{6}}=\left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$ $AM \geq GM$ $\displaystyle \frac{1}{6} \geq \left( \frac{ab^2 c^3}{2^2 3^3}\right)^{1 \over 6}$ $\displaystyle \Rightarrow \frac{2^23^3}{6^6} \geq ab^2c^3$