Is it possible to prove that for $x,y,z \in \Bbb{R}$; if $x,y,z >0$ and $xyz=1$, then $(x+y+z)\ge 3$ without using the AM-GM inequality?

Question

I'm asked in an exercise from an algebra textbook to prove that for$\{ x,y,z\} \subset \Bbb{R}$; if $x,y,z >0$ and $xyz=1$, then $x+y+z\ge 3$.

Using the arithmetic and geometric mean inequality the proof is easy. But in this book the definitions of roots, logarithms and inductive sets are introduced in later chapters. So I guess there is a way to prove it without AM-GM because most proofs of the AM-GM inequality that I'm aware of requiere either induction or logarithms and the inequality itself involves a root.

Can anyone please help me? Is there a proof without using AM-GM?

Answer 1

The proof without AM-GM.

Let $x=a^3$, $y=b^3$ and $z=c^3$.

Hence, $abc=1$ and $$x+y+z-3=a^3+b^3+c^3-3=a^3+b^3+c^3-3abc=$$ $$=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=$$ $$=\frac{1}{2}(a+b+c)((a-b)^2+(a-c)^2+(b-c)^2)\geq0.$$ Done!

Answer 2

Another way.

We need to prove that $$a^3+b^3+c^3-3abc\geq0.$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, our inequality is a linear inequality of $v^2$ because it's third degree.

Thus, it remains to prove our inequality for an extremal value of $v^2$,

which happens for equality case of two variables.

Since the last inequality is homogeneous and symmetric, we can assume $v=c=1$,

which gives $$a^3+2-3a\geq0$$ or $$(a+2)(a-1)^2\geq0.$$ Done!

Answer 3

Also, we can use SOS here: $$a^3+b^3+c^3-3abc=\sum_{cyc}(a^3-abc)=\sum_{cyc}a(a^2-bc)=$$ $$=\frac{1}{2}\sum_{cyc}a((a-b)(a+c)-(c-a)(a+b))=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)(a(a+c)-b(b+c))=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)(a^2-b^2+ac-bc)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b+c)\geq0.$$