An algebra problem by I Gede Arya Raditya Parameswara

Algebra Level 4

Let x , y x,y and z z be positive numbers satisfying 1 x + 1 y + 1 z = 6 \dfrac1x+\dfrac1y + \dfrac1z = 6 . What is the maximum value of 1 x 3 y 2 z \dfrac1{x^3 y^2 z} ?

Give your answer to 2 decimal places.


The answer is 108.00.

I Gede Arya Raditya Parameswara by I Gede Arya Raditya Para… , 17

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rishabh Jain
Dec 25, 2016

Applying G M ≥ H M GM\ge HM on 3 x , 3 x , 3 x , 2 y , 2 y , z 3x,3x,3x,2y,2y,z :

108 x 3 y 2 z 6 ≥ 6 1 3 x + 1 3 x + 1 3 x + 1 2 y + 1 2 y + 1 z ⏟ 1 x + 1 y + 1 z = 6 \sqrt[6]{108x^3y^2z}\ge\dfrac{6}{\underbrace{\frac{1}{3x}+\frac{1}{3x}+\frac{1}{3x}+\frac{1}{2y}+\frac{1}{2y}+\frac{1}{z}}_{\frac{1}x+\frac{1}y+\frac 1z=6}}

⟹ 108 x 3 y 2 z ≥ 1 \implies 108x^3y^2z\ge 1

⟹ 1 x 3 y 2 z ≤ 108 \implies \dfrac{1}{x^3y^2z}\le \boxed{108}

Equality occurs when 3 x = 2 y = z 3x=2y=z or x = 1 3 , y = 1 2 , z = 1 x=\dfrac 13,y=\dfrac 12,z=1 .

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Why not am Gm?

I Gede Arya Raditya Parameswara - 4 years, 5 months ago

Put 1/x=a ,1/y=b ,1/z=c then u can easily use AM-GM

Kushal Bose - 4 years, 5 months ago

Log in to reply

Without putting also, AM-GM can be applied directly but introducing HM is only a convenient method to apply it...

Rishabh Jain - 4 years, 5 months ago

Log in to reply

Yes but this will be easy to handle

Kushal Bose - 4 years, 5 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...