I'm reading some exercises from Cvetovski's "Inequalities", 2012, that seem to imply that it is safe to multiply both sides of an inequality by a real variable. For example, proving $\forall x \in \mathbb{R}: x + 1/x >= 2$ like so

$$ (x-1)^2 >= 0 \\ x^2 - 2x + 1 >= 0 \\ x^2 + 1 >= 2x \\ x + 1/x >= 2 $$

This last step appears to say that it is safe to multiply both sides of an inequality by a real factor and to maintain the same directionality of the inequality.

However, intuition could lead some to conclude that it is not correct to do this, as the unknown, lacking restrictions other than $\in \mathbb{R}$, could be negative, in which case

$$ 3 > 2 \\ 3 \cdot -1 = -3 \\ 2 \cdot -1 = -1 \\ -3 < -1 \\ 3 \cdot -1 < 2 \cdot -1 $$

which inverts the relation.

Is the book completely wrong, temporarily glossing over a rule it will announce later, or have I made a mistake?

Also, what about zero? Is multiplying by zero allowed for -or-equals ($\le, \ge$) relations, but not for the -but-not-equals ($<, >$)?

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    $\begingroup$ You have to break it into cases because $x$ need not be positive. Only multiplication/division by positives preserves inequalities (as you note). Note that if $x < 0$, so is $\frac{1}{x}$ so the inequality does not hold for $x<0$ (and is undefined for $x=0$). $\endgroup$ Jun 30, 2016 at 1:53
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    $\begingroup$ It is worth mentioning that $x+\frac{1}{x}$ is negative when $x$ is negative, and could not possibly be greater than two. The inequality is true when $x$ is a positive real number. It is also worth mentioning that dividing by zero is not allowed and so care needs to be used in that case as well. $\endgroup$
    – JMoravitz
    Jun 30, 2016 at 1:55
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    $\begingroup$ SORRY, but $2\cdot (-1)=\color{red}{-2}$ $\endgroup$ Jun 30, 2016 at 2:03
  • $\begingroup$ This is only acceptable if you know x>0. Then it is always acceptable. $\endgroup$
    – fleablood
    Jun 30, 2016 at 2:54

2 Answers 2


It's true that, for all positive reals, we have $x + \frac{1}{x} \geq 2$, but try any negative real value for $x$, and you will see that the claim does not hold for negatives. The proof is correct if we are making the assumption that $x>0$.

To answer your general question, multiplying inequalities by positive numbers preserves the relations, and multiplying by negative numbers reverses it, as you have observed. Multiplying by zero takes away all information present in the inequality, and you're right that it ruins strict inequalities to do that.

Multiplying an inequality by a variable expression, you have to split into cases: one case where the variable expression is $>0$ and the relation remains the same, and another case where the variables expression is $<0$ and the relation reverses, and a special case if the variable expression $=0$, which is generally easy to handle. In this problem, you needn't consider the third case, because having $\frac{1}{x}$ in the original expression removes $x=0$ from consideration right off the bat.

What's really going on here is that inequalities are preserved by increasing functions, and reversed by decreasing functions - this follows directly from the definition of an increasing/decreasing function. Multiplication by a positive number corresponds to the increasing function $f(x)=kx$ with positive $k$. Multiplication by $0$ corresponds to plugging all expressions into the constant function $f(x)=0$. After doing that, the relation between $f(x_1)$ and $f(x_2)$ tells you nothing about the original relation between $x_1$ and $x_2$.


Re-reading the textbook problem description one last time, in fact we see the additional information $x>0$. Oops.


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