1. ## AMGM Inequality help?

Alright, here goes.

Prove that
abc > (a+b-c)(a-b+c)(-a+b+c)

What I've done:

By AM>GM property,
(a+b+c)/3 > (abc)^(1/3)

By AM>GM property, I also found that
(a+b+c)/3 > ((a+b-c)(a-b+c)(-a+b+c))^(1/3)

From here onwards I'm stuck, since it doesn't really make sense now.
Need help here.

2. Let $\displaystyle x=a+b-c, \ y=a-b+c, \ z=-a+b+c$

Then $\displaystyle a=\frac{x+y}{2}, \ b=\frac{x+z}{2}, \ c=\frac{y+z}{2}$

and the inequality becomes $\displaystyle (x+y)(x+z)(y+z)\geq 8xyz$

Now apply three times $\displaystyle AM\geq GM$.

3. Originally Posted by red_dog
Now apply three times $\displaystyle AM\geq GM$.
$\displaystyle (x+y)(x+z)(y+z)\geq 8xyz$

I've pondered for a while.
But ultimately I get this:

Applying $\displaystyle AM\geq GM$,

$\displaystyle \frac{(2x+2y+2z)^3}{27} \geq (x+y)(x+z)(y+z) \geq 8xyz$

Sorry if I am abit slow on understanding.

4. $\displaystyle x+y\geq 2\sqrt{xy}$

$\displaystyle x+z\geq 2\sqrt{xz}$

$\displaystyle y+z\geq 2\sqrt{yz}$

Then, $\displaystyle (x+y)(x+z)(y+z)\geq 8\sqrt{x^2y^2z^2}=8xyz$

5. My deepest gratitude for offering your solution.