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.

John Raymond Diaz

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

 

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