How can I prove this inequality?

$$\frac{{1+2x}}{1+x}\geq \frac{1+2y}{1+y}$$

  • 5
    $\begingroup$ Answers > votes > work shown $\endgroup$ – The Chaz 2.0 Apr 26 '12 at 23:48

10 Answers 10


$$ \frac{1+2x}{1+x} - \frac{1+2y}{1+y} = \frac{(1+y)(1+2x) - (1+x)(1+2y)}{(1+x)(1+y)} = \frac{x-y}{(1+x)(1+y)}$$


Cross-multiply: The statement given is true if and only if:$$(1+2x)(1+y)\geq (1+2y)(1+x)$$ if and only if$$1+2x+y+2xy\geq 1+2y+x+2xy$$ if and only if $$x \geq y$$

  • $\begingroup$ The first step uses the postulate, that rather breaks "if and only if". $\endgroup$ – Ben Voigt Apr 26 '12 at 23:54
  • $\begingroup$ @BenVoigt: Huh? The first statement holds if and only if the last statement holds. How are you "breaking" the chain? $\endgroup$ – Arturo Magidin Apr 27 '12 at 4:10
  • $\begingroup$ @Arturo: The cross-multiplication step relies on the OTHER part of the postulate, that x+1 and y+1 are both positive. $\endgroup$ – Ben Voigt Apr 27 '12 at 4:32
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    $\begingroup$ @BenVoigt: It is all being done under the assumption of $x,y$ positive, yes. $\endgroup$ – Arturo Magidin Apr 27 '12 at 4:33

The function $$ f(x)=\frac{1+2x}{1+x}=1+\frac{x}{1+x}=2-\frac{1}{1+x} $$ is a strictly increasing function, i.e., $x > y \implies f(x) > f(y)$. This can be seen if you graph it, take its derivative if calculus is allowed or perhaps simplest of all, if you notice, after André Nicolas's kind comment, that $x > y > 0$ $\implies 1+x > 1+y$ $\implies \frac1{1+x} < \frac1{1+y}$ $\implies -\frac1{1+x} > -\frac1{1+y}$ $\implies 2-\frac1{1+x} > 2-\frac1{1+y}$ (when we take the reciprocal of both sides, we need that $1+x$ and $1+y$ have the same sign).

Its graph looks like $y=-\frac1x$ but with the origin shifted to $(-1,2)$ (and then appropriately clipped for $x > 0$), as can be seen by substituting $y-2$ for $y$ and $x+1$ for $x$ in the above. In particular, the graph on the whole real line consists of two hyperbolae in the second and fourth quadrants and horizontal and vertical asymptotes along the axes, all relative to the shifted origin.

Its derivative is $f\,'(x)=(x+1)^{-2}$ which is strictly positive for $x\ne-1$.

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    $\begingroup$ Good mention of the geometry. At the algebra level, if $x>y$, then $2-\frac{1}{1+x}$ subtracts less from $2$ than $2-\frac{1}{1+y}$. $\endgroup$ – André Nicolas Apr 26 '12 at 22:29

As $(1+x)(1+y)$ is positive, we have

\begin{eqnarray*} & \dfrac{1+ 2x}{1+x} & \ > & \dfrac{1 +2y }{1 + y} \\ \Leftrightarrow & (1+ 2x)(1+y) & \ > & (1+2y)(1+x) \\ \Leftrightarrow & 1 + 2x + y + 2xy &\ >& 1+ 2y + x + 2yx \\ \Leftrightarrow & x & \ > & y \\ \end{eqnarray*}

  • $\begingroup$ You should have $\geq$, not $>$. $\endgroup$ – Pedro Tamaroff Apr 26 '12 at 22:58

Using trigonometry:

Put $x = \tan^2{\theta}$ for $\theta \in (0, \frac{\pi}{2})$.

$$\frac{1 + 2x}{1 +x} = \frac{1 + 2 \tan^2 \theta}{\sec^2 \theta} = \cos^2 \theta + 2 \sin^2 \theta = 1 + \sin^2 \theta$$

Since $\sin^2 \theta$ and $\tan^2 \theta$ are strictly increasing functions of $\theta \in (0, \frac{\pi}{2})$, we are done.


It's sometimes nice to work backward, as long as you can reverse each step you take--for example, don't multiply an equation by zero (not that you would), because there's no going back from that.

Note that since $x>y>0$, then $1+x>1+y$, and $1+x,1+y>0$ (so multiplication or division by $1+x,1+y$ won't have any effect on inequalities, and is reversible). Further note that for any real $z$, we have $1+2z=1+z+z$, so as long as $z\neq-1$, $\frac{1+2z}{1+z}=\frac{1+z}{1+z}+\frac{z}{1+z}=1+\frac{z}{1+z}$. With these observations, we can work backward from the desired conclusion (as expressed in the title of the post). In particular, given any real $x,y>0$, the following steps all work out. $$\begin{eqnarray*} \frac{1+2x}{1+x}>\frac{1+2y}{1+y} & \Leftrightarrow & 1+\frac{x}{1+x}>1+\frac{y}{1+y}\\ & \Leftrightarrow & \frac{x}{1+x}>\frac{y}{1+y}\\ & \Leftrightarrow & \frac{x}{1+x}(1+x)(1+y)>\frac{y}{1+y}(1+x)(1+y)\\ & \Leftrightarrow & x+xy>y+xy\\ & \Leftrightarrow & x>y.\end{eqnarray*}$$

Once you've got something like this, you can provide a nice, step-by-step progression from start to finish. It isn't always easy to work from end-to-beginning this way, nor is a beginning-to-end approach always best. Sometimes you may want to work from beginning and end, and try to make them meet in the middle. It's good to stay flexible.


It's good, if possible, to try to understand why the inequality is true before embarking on algebraic manipulations. Let's try an example: $3\gt 2$ and $$ \dfrac{1+2\cdot3}{1+3}=1\!\frac{3}{4}\gt1\!\frac{2}{3}=\dfrac{1+2\cdot2}{1+2}. $$ Notice the pattern in the mixed fractions in the middle. The general idea is that $\dfrac{1+2x}{1+x}$ reduces to the number $1+\dfrac{x}{1+x}$. You should prove this algebraically.

You can now see that the term $1$ in $1+\dfrac{x}{1+x}$ is irrelevant to the inequality: the original inequality is true because $\dfrac{x}{1+x}\gt\dfrac{y}{1+y}$ when $x\gt y$. To understand why this new inequality is true, notice that numbers of the form $\dfrac{x}{1+x}$ are numbers like $\frac{3}{4}$, $\frac{4}{5}$, $\frac{5}{6}$, $\ldots$. As the numerator gets bigger, the fraction gets bigger. To prove this formally, observe that such numbers could also be written $1-\dfrac{1}{x}$. As $x$ gets bigger, $\dfrac{1}{x}$ gets smaller, and so $1-\dfrac{1}{x}$ gets bigger. You should now try to turn this into an algebraic proof.


The following statements are equivalent

  • $+\infty \gt x \gt y \gt 0$

  • $+\infty \gt 1+x \gt 1+y \gt 1$

  • $0 \lt \dfrac{1}{1+x} \lt \dfrac{1}{1+y} \lt 1$

  • $0 \gt -\dfrac{1}{1+x} \gt -\dfrac{1}{1+y} \gt -1 $

  • $2 \gt 2-\dfrac{1}{1+x} \gt 2-\dfrac{1}{1+y} \gt 1 $

  • $2 \gt \dfrac{1+2x}{1+x} \gt \dfrac{1+2y}{1+y} \gt 1 $







  • 1
    $\begingroup$ Just for the benefit of future visitors, start with $$x > y$$ Add $1 + x + y + 2xy$ to both sides. Factor to get $$(1+2x)(1+y)>(1+2y)(1+x)$$ Divide through by $(1 + y)(1 + x) > 0$ to get the result. $\endgroup$ – The Chaz 2.0 May 3 '12 at 19:27
  • $\begingroup$ And my strategy was "work backwards" as Cameron Buie mentioned. After cross multiplying out and arriving at $x>y$, the steps are just reversible. I wrote it extremely mysteriously, but if you approach it this way you will get the same results without trouble. $\endgroup$ – rschwieb May 3 '12 at 19:29

Limit-Based Answer bridging into Calculus

Firstly, one opening remark: for consistency, always write polynomials in descending degree: $2x + 1$ not $1 + 2x$.

The function $f(x) = {{2x + 1}\over{x + 1}}$ monotonically increases for positive x. To show this, take some arbitrary positive $a$, and show that $\lim_{h\to 0}{{f(a + h) - f(a)}\over h}$ is positive.

This particular limit formula, $\lim_{h\to 0}{{f(a + h) - f(a)}\over h}$, leads to the calculus concept of a derivative, providing a bridge from precalculus to calculus.

What the $\lim_{h\to 0}$ notation says is that we have a limit parameter called $h$ and we are making it arbitrarily close to zero, to see what effect this has on the formula which follows.

Our formula is structured such that it uses this small number $h$ to make a step in the function: go from the domain value $a$ to $a + h$, and divide the tiny change in the function's value $f(a+h) - f(a)$ by the step size $h$. But you will probably recognize that this just the calculation of the slope: rise divided by run!

If we can show that the slope of the function is positive at any $a > 0$, it means that the function $f(x)$ increases for all positive $x$.

If you know calculus you can just work out the derivative of $f$, by using the rules and trick for computing derivatives, and show that it is positive for $a > 0$.

Not yet knowing calculus, we can work in detail through through the limit calculation:

$$\lim_{h\to 0}{{f(a + h) - f(a)}\over h}$$

$$\lim_{h\to 0}{{{{2(a + h) + 1}\over{a + h + 1}} - {{2a + 1}\over{a + 1}}}\over h}$$

$$\lim_{h\to 0}{{{{(a + 1)(2(a + h) + 1) - (2a + 1)(a + h + 1)}\over{(a + h + 1)(a + 1)}}}\over h}$$

When you expand the topmost numerator, you will find that most of the terms cancel to leave just $h$:

$$\lim_{h\to 0}{{{{h}\over{(a + h + 1)(a + 1)}}}\over h}$$

And this cancels with the bottom h:

$$\lim_{h\to 0}{{1\over{(a + h + 1)(a + 1)}}}$$

Long before arriving at the limit, which in the very next step we obtain by setting $h = 0$, we can already see that this limit is positive for $a > 0$:

$${{1\over{(a + 1)(a + 1)}}}$$

Since $\lim_{h\to 0}{{f(a + h) - f(a)}\over h} > 0$ it means that $f$ increases.

(Yes, I did say that $h$ never reaches zero; but in computing limits, there are situations when it's both convenient and justified for the limit parameter to vanish. E.g. the limit of $x + a$ as $a$ approaches zero is just $x$. Well, so arbitrarily close to $x$ that the difference between that limit and $x$ is negligible!)


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