Prove $x+y+z\geq3$ given $xyz=1$ and $0

Question

Prove $x+y+z\geq3$ given $xyz=1$ and $0<x,y,z\in\mathbb{F}$. Secondly, show that $x+y+z=3\iff x=y=z=1$.

($\mathbb{F}$ is the ordered field)

I've tried applying the following lemma with no success yet:

If $0<a<1$ and $1<b$ where $a,b\in\mathbb{F}$ then $1+ab<a+b$

is there a reason why you specifically want to use the lemma? theres only a handful of ways in which $xyz=1$ even with the inclusion of complex numbers. — Vaas, Nov 02 '17 at 12:33
@Vaas thought it could help as it was previously proven (one problem earlier) — Slavik Egorov, Nov 02 '17 at 12:44
This answer proves the more general result, your question is the case $n=3.$ — bof, Nov 02 '17 at 12:45
I have no idea about field. Can I do the problem using Calculus?(Multiplier). It is easier. — , Nov 02 '17 at 14:03

Kim Jong Un · Accepted Answer · 2017-11-05T15:47:11.443

3

Let $D=(x+y+z)^3-27xyz$ and $S=x+y+z$. Then $$ 2D=(S+6z)(x-y)^2+(S+6x)(y-z)^2+(S+6y)(z-x)^2\geq 0 $$ and so $(x+y+z)^3\geq 27xyz=27$ and therefore $x + y + z\geq 3$. We have $x + y +z=3$ iff $D=0$ iff $x=y=z$.

edited Nov 05 '17 at 15:47

answered Nov 02 '17 at 12:59

Kim Jong Un

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I found this proof, but you beat me. +1. – Michael Rozenberg Nov 02 '17 at 13:11
You may need to show $a^3\ge b^3\implies a\ge b$ using ordered field axioms for completeness but that isn’t hard. +1 – Macavity Nov 02 '17 at 13:15
Why is this mandatory? $$ 2D=\sum_{\text{cyc}}(x+y+7z)(x-y)^2\geq 0 $$ – Slavik Egorov Nov 02 '17 at 14:45
@SlavikEgorov The equality holds by direct verification and the inequality because $x,y,z>0$. – Kim Jong Un Nov 02 '17 at 17:07
Could you explain that cyc saying under the sigma and what did you apply to get that expression? – Slavik Egorov Nov 05 '17 at 13:38
@SlavikEgorov I have rewritten the result without the cyclical summation notation. To verify the identity, just expand both sides. – Kim Jong Un Nov 05 '17 at 15:47
Also, how do you prove $ x+y+z =3\iff x=y=z=1 $ ? – Slavik Egorov Nov 06 '17 at 19:08
@SlavikEgorov In $$2D=(S+6z)(x-y)^2+(S+6x)(y-z)^2+(S+6y)(z-x)^2,$$ $x=y=z$ is the only way to make the RHS $0$. – Kim Jong Un Nov 06 '17 at 19:33

score 0 · Answer 2 · answered Nov 02 '17 at 15:22

Let $f(x,y,z)=x+y+z$, $g(x,y,z)=xyz$

Solve the equations $\bigtriangledown f=\lambda \bigtriangledown g \tag {1}$

$g(x,y,z)=1 \tag{2}$

$x, y, z \ge 0$

$\implies$ $1=\lambda yz\tag{*}$ $1=\lambda xz\tag{**}$ $1=\lambda xy\tag{***}$ $g(x,y,z)=1 \tag{****}$ $\implies$ $x=y=z=\lambda$

Appliying in (2) we get the desired result.

Prove $x+y+z\geq3$ given $xyz=1$ and $0

2 Answers2