How to prove (x+y)/z + (y+z)/x + (x+z)/y>=6 when x,y,z>0

Question

This question actually had two parts. In the first part I had to prove that a + 1/a >=2. I proved it by rearranging it to (a-1)^2 >= 0, which is always true.

So, I thought the second problem would require a similar method.

(x+y)/z + (y+z)/x + (x+z)/y >=6, where x,y,z>0

But I cant figure it out. I've tried simplifying it and factoring it for ideas but I've got nothing.

I'm voting to close this question as off-topic because is not programming question — eyllanesc, Feb 25 '19 at 20:23
I'm voting to close this question as off-topic because it is about mathematics instead of directly about programming / coding / programming tools / software algorithms. — Pang, Feb 28 '19 at 02:09

score 0 · Accepted Answer · answered Feb 25 '19 at 19:43

Once you know that a + 1/a >= 2, the second part is easy. Define:

a := x/z,  b := y/z,  c := y/x

and now

(x+y)/z + (y+z)/x + (x+z)/y = x/z + y/z + y/x + z/x + x/y + z/y
                            = a + b + c + 1/a + 1/c + 1/b
                           >= 2 + 2 + 2
                            = 6

How to prove (x+y)/z + (y+z)/x + (x+z)/y>=6 when x,y,z>0

1 Answers1