# Kobe’s AM-GM Inequality

Today’s featured problem is from a 16-year old and incoming freshman student at Pamantasan ng Lungsod ng Maynila named Kobe Bryant L. Pe. He graduated in Caloocan High School. He is not an NBA superstar. He is a Math-star. His mathematics career started when he was in 1st year and started winning in math contest when he was in the second year of his high school life. Here are some of his achievements.

2012 MMC camanava orals- 2nd placer

2012 MMC sectoral finalist

2013 MMC camanava orals -3rd placer

2013 MMC sectoral finalist

2013 MMC regional finalist

MCU Interschool- team- 2nd placer

The following is the problem he wanted to share.

Problem:

Suppose that a,b,c are non-negative real numbers,  find the minimum possible value of

$\displaystyle\frac{a^2b+ac^2+8b^2c}{5abc}$

Solution:

As we can see, we can split the expression into three fractions:

$\displaystyle\frac{a^2b}{5abc}+\displaystyle\frac{ac^2}{5abc}+\displaystyle\frac{8b^2c}{5abc}$

Simplifying,

$\displaystyle\frac{a}{5c}+\displaystyle\frac{c}{5b}+\displaystyle\frac{8b}{5a}$

By AM-GM inequality,

$\displaystyle\frac{a}{5c}+\displaystyle\frac{c}{5b}+\displaystyle\frac{8b}{5a}\geq 3[\root 3 \of{\frac{a}{5c}\cdot \frac{c}{5b}\cdot \frac{8b}{5a}}]$

$\displaystyle\frac{a}{5c}+\displaystyle\frac{c}{5b}+\displaystyle\frac{8b}{5a} \geq 3[\root 3 \of{\frac{8}{125}}]$

$\displaystyle\frac{a}{5c}+\displaystyle\frac{c}{5b}+\displaystyle\frac{8b}{5a} \geq 3\cdot \frac{2}{5}$

$\displaystyle\frac{a}{5c}+\displaystyle\frac{c}{5b}+\displaystyle\frac{8b}{5a} \geq \displaystyle\frac{6}{5}$

Since,

$\displaystyle\frac{a^2b}{5abc}+\displaystyle\frac{ac^2}{5abc}+\displaystyle\frac{8b^2c}{5abc}=\displaystyle\frac{a^2b+ac^2+8b^2c}{5abc}$

So,

$\displaystyle\frac{a^2b}{5abc}+\displaystyle\frac{ac^2}{5abc}+\displaystyle\frac{8b^2c}{5abc}\geq \displaystyle\frac{6}{5}$

Therefore, the minimum possible value is   $\displaystyle\frac{6}{5}$