how to solve this inequiality $xyz ⩾ (x+y-z)(y+z-x)(z+x-y)$

Question

Assuming positive real values of $x,y,z$, how can we prove that $(xyz) ⩾ (x+y-z)(y+z-x)(z+x-y)$

That should be a $\geqslant$ instead of > – iamwhoiam Jul 17 '17 at 04:35 — iamwhoiam, Jul 17 '17 at 04:35

score 1 · Accepted Answer · answered Jul 17 '17 at 04:35

1

hint: $(a+b-c)(b+c-a) \le b^2$, by AM-GM. Use this pairwise for $3$ pairs. You can assume $a,b,c$ are sides of a triangle for if not the inequality is trivially true.

answered Jul 17 '17 at 04:35

DeepSea

77,651

Got it. Thanks a lot for hint. – Chandan Dey Jul 17 '17 at 04:43
It’s really an awesome hint. What I want to know is how you got it? I mean in Algebra how do we come up with those things? – Knight wants Loong back Aug 11 '20 at 09:50

how to solve this inequiality $xyz ⩾ (x+y-z)(y+z-x)(z+x-y)$

1 Answers1

Linked