An algebra problem by Mardokay Mosazghi

Algebra Level 1

If xyz = 27, x, y, and z are positive, find the minimum value of x + y + z.

2 solutions

Daniel Liu
Jun 23, 2014

By AM-GM, $\dfrac{x+y+z}{3}\ge \sqrt[3]{xyz}=3$

Thus, $x+y+z\ge \boxed{9}$

I was just about to write the same solution! Why'd you beat me by 2 minutes.

Sharky Kesa - 6 years, 11 months ago

^_^ I wrote my solution as fast as possible.

Daniel Liu - 6 years, 11 months ago

@daniel Liu You are amazing, you always post solutions faster than me even on my own problems, one day i strive to beat you.

Mardokay Mosazghi - 6 years, 11 months ago

Why's the question called Vieta easy?

Victor Loh - 6 years, 11 months ago

H ha I didnot think about it i was doing the vieta problems by calvin and i guess i just typed it in unknowingly.

Mardokay Mosazghi - 6 years, 11 months ago

Actually is there a way to approach this problem by means of Vieta's equations?

Victor Loh - 6 years, 11 months ago

@Victor Loh – I am going to try it

Mardokay Mosazghi - 6 years, 11 months ago

Mardokay Mosazghi, you are exceptional

Carlos Alejandro Palma Bernal - 6 years, 10 months ago

Question is wrong

Naheel Ali - 6 years, 11 months ago

@Naheel Ali Why is the question wrong?

Mardokay Mosazghi - 6 years, 11 months ago

Grace Fernandez
Aug 3, 2014

27 divided by 3 is equals to 9