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I'm working through some proofs from Cvetkovski's "Inequalities", when I came across a more difficult one (for newbies like me). Given $a, b, c \in \mathbb{R} \mid a+b+c \geq abc$, how can we prove that $3abc(a+b+c) \geq 3(abc)^2$ ?

On the surface, it appears that we could do something like:

$$ \begin{align*}abc &\geq abc \\ abc(a+b+c) &\geq abc(abc) \\ abc(a+b+c) &\geq (abc)^2 \\ 3abc(a+b+c) &\geq 3(abc)^2\end{align*} $$

However, the second line is no good, as inequalities, unlike equations, cannot be safely multiplied by quantities of unknown sign, because in the case that the multiplier is negative, the sign must change, whereas in the case that the multiplier is positive, the sign must maintain its direction.

I can think of a few other ways to accomplish the proof, like squaring both sides of $a+b+c \geq abc$ instead of multiplying both sides by $abc$; or by exhaustion of the positive/negative possibilities of $abc$ and $a+b+c$, but these strategies appear (to my novice hands) to not be very fruitful.

Any tips for completing this proof?

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Let $a = -1, b = 2, c = 2$. The left side is negative but the right is positive. Are you sure it's not $a, b, c \in \mathbb{R}^+$? You're right about being hesitant about multiplying by $abc$ on both sides.

With the added restriction that $a, b, c \in \mathbb{R}^+$, or even more generally that $abc \geq 0$, your method works fine.

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  • $\begingroup$ Agreed. The "Inequalities" 2012 edition mostly likely has a typo in the constraint ($\mathbb{R}$ vs. $\mathbb{R+}$), as most of the earlier problems are constrained to the positive reals. $\endgroup$ – mcandre Jul 6 '16 at 19:11
  • $\begingroup$ Er, "Inequalities" exercise 1.8 actually does work for negative reals, as $\forall a,b,c \in \mathbb{R^-} \mid a+b+c \geq abc: a^2+b^2+c^2$ is trivially $\geq abc \sqrt3$ $\endgroup$ – mcandre Jul 6 '16 at 21:38
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If $A \ge B$, then $C A \ge CB$ if and only if ($C \ge 0$ or $A = B$). So, as Michael Tong noted, you do want $abc \ge 0$.

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