# Raymond’s Simple AM-GM inequality Trick

Featured problem for today is from Raymond John Diaz, another Math quizzer and school heartthrob. He is a graduate of Pedro Guevara Memorial National High School. Currently, he is taking up BS Applied Mathematics at University of the Philippines – Los Baňos (UP-LB).

He first submitted a problem about geometry but we both agreed to change the problem to a different one since that problem was quite vague. He says he wants to pursue Actuary-the highest paying professionals. Raymond is on his 4th level in algebra in brilliant.

Here are some of his achievements

– 2011, 2012 and 2014 MMC Regional Finals (team) 1st Runner up

-15th and 16th PMO Area Stage qualifier -Southern Tagalog

-Southern Tagalog Invitational Mathematical Challenge a.k.a Mathematch (Team 1st runner up)(individual Top scorer written phase)

-MCL Cup Math Wizards (team champion)

Here is the problem he wants to share about am-gm inequality.

Problem:

Let $a,b,c,d >0$
Show that $(a+b+c+d)(\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}+\displaystyle\frac{1}{d})\ge 16$

Solution: $(a+b+c+d)(\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}+\displaystyle\frac{1}{c}+\displaystyle\frac{1}{d})\ge 16$

First we need to multiply the two expressions at the left side of the equation $1+\displaystyle\frac{a}{b}+\displaystyle\frac{a}{c}+\displaystyle\frac{a}{d}+\displaystyle\frac{b}{a}+1+\displaystyle\frac{b}{c}+\displaystyle\frac{b}{d}+\displaystyle\frac{c}{a}+\displaystyle\frac{c}{b}+1+\displaystyle\frac{c}{d}+\displaystyle\frac{d}{a}+\displaystyle\frac{d}{b}+\displaystyle\frac{d}{c}+1\ge 16$

Then we need to group it as sum of their reciprocals and subtract 4 from both sides $(\displaystyle\frac{a}{b}+\displaystyle\frac{b}{a})+(\displaystyle\frac{a}{c}+\displaystyle\frac{c}{a})+(\displaystyle\frac{b}{c}+\displaystyle\frac{c}{b})+(\displaystyle\frac{b}{d}+\displaystyle\frac{d}{b})+(\displaystyle\frac{c}{d}+\displaystyle\frac{d}{c})\ge 12$

Now we let $\displaystyle\frac{a}{b}+\displaystyle\frac{b}{a}=\displaystyle\frac{a^2+b^2}{ab}$       (i) $\displaystyle\frac{a}{c}+\displaystyle\frac{c}{a}=\displaystyle\frac{a^2+c^2}{ac}$       (ii) $\displaystyle\frac{a}{d}+\displaystyle\frac{d}{a}=\displaystyle\frac{a^2+d^2}{ad}$       (iii) $\displaystyle\frac{b}{c}+\displaystyle\frac{c}{b}=\displaystyle\frac{b^2+c^2}{bc}$       (iv) $\displaystyle\frac{b}{d}+\displaystyle\frac{d}{b}=\displaystyle\frac{b^2+d^2}{db}$       (v) $\displaystyle\frac{c}{d}+\displaystyle\frac{d}{c}=\displaystyle\frac{c^2+d^2}{cd}$       (vi)

By the AM-GM inequality for $x,y>0$ , we have the lowest value of $\displaystyle\frac{x^2+y^2}{xy}=2$

Proof: $\displaystyle\frac{x+y}{2}\ge\sqrt{xy}$

Square both sides $\displaystyle\frac{x^2+2xy+y^2}{4}\ge xy$

Multiply both sides by $\displaystyle\frac{4}{xy}$ $\displaystyle\frac{x^2+2xy+y^2}{xy}\ge 4$ $\displaystyle\frac{x^2+y^2}{xy}\ge 2$

Therefore the lowest value of (i) + (ii) + (iii) + (iv) + (v) + (vi) equal to $12$