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$a, b, c$ 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.

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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.

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