Prove inequality of natural x, y, z

Question

Prove that, for every x , y, z of the natural inequality is true:

$x\ $ $+$ $y\ $ $+$ $z\ $ $\ge$ $3$ $and$ $x$*$y$*$z$ $=$ $1$

I tried to divide by 3, but does not go :(

I'm not sure I understand this correctly. you want to proove that vor every x,y,z ($\in\mathbb{R}$?) for which $xyz=1$ is true $x+y+z \geq 3$? — Börge, Oct 14 '15 at 16:31

score 2 · Answer 1 · answered Oct 14 '15 at 16:46

2

HINT: we have $\frac{x+y+z}{3}\geq \sqrt[3]{xyz}=1$ by AM-GM

answered Oct 14 '15 at 16:46

Where does the root level 3? – JWa Oct 14 '15 at 16:50
if you mean what it is, it is the revert function to f(x) = x^3 – Börge Oct 14 '15 at 16:51
Ok, I know now :) – JWa Oct 14 '15 at 16:51
can you use the AM-GM? – Börge Oct 14 '15 at 16:51
No :( I can't :( – JWa Oct 14 '15 at 16:52
yes AM-GM works for numbers $a\geq 0,b\geq 0,c\geq 0$ – Dr. Sonnhard Graubner Oct 14 '15 at 16:53
What's next? This is end of prove? – JWa Oct 14 '15 at 16:54
yes this is the end of the proof – Dr. Sonnhard Graubner Oct 14 '15 at 16:56
I ment: are you allowed to use the AM-GM..... – Börge Oct 14 '15 at 16:57
if it is not allowed he must remark ist – Dr. Sonnhard Graubner Oct 14 '15 at 16:58
Yes, you can use this :) Sorry – JWa Oct 14 '15 at 16:58
I solve this ;) thanks – JWa Oct 14 '15 at 21:34

1 Answers1