# AM-GM inequality basic

$a, b, c$ are positive real numbers.

Prove that $$8abc\le (a+b)(b+c)(c+a)$$

I haven’t really gotten anywhere, just tried to open up $(a+b)(b+c)(c+a)$ and got null.

#### Solutions Collecting From Web of "AM-GM inequality basic"

It’s really as simple as noting that $a+b\geq 2\sqrt{ab}$, and similarly for $a+c$ and $b+c$, then multiplying it out.

Using AM-GM, we get that
$$\frac{a+b}{2} \geq \sqrt{ab}$$
$$\Rightarrow a+b \geq 2 \sqrt{ab}$$

Similar manipulations show that
$$b+c \geq 2 \sqrt{bc}$$
$$a+c \geq 2 \sqrt{ac}$$

Multiplying all these inequalities together ( this is allowed here, since all numbers involved are positive), we get
\begin{align} (a+b)(b+c)(a+c) &\geq (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ac}) \\ &= 8 \sqrt{ab.bc.ac} \\ &= 8 \sqrt{a^{2} b^{2} c^{2}} \\ &= 8abc \end{align}
as required.