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Prove that if $x > y > 0,$ then $\sqrt{xy} < (x-y)/(\ln x - \ln y) < (x+y)/2$.

I know that the area under the hyperbola $s = \frac{1}t$ from $t = y$ to $x$ is $\ln x - \ln y$. I also know that the area of the trapezoid formed by the lines $t=x,y$, the x-axis, and a tangent to this hyperbola between x and y is less than $\ln x - \ln y$ by convexity. Given a point $(t_0,s_0)$ on the hyperbola, the tangent has equation $s - s_0 = -\frac{1}{t_0^2}(t-t_0) $. But how can I find some points $(t_0, s_0)$ that might be useful for solving this problem?

user3472
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2 Answers2

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There are easier ways to find the area of a right trapezoid. If it has 'base' length $b$ and 'side' lengths $a_1$ and $a_2$, its area is $\frac{a_1+a_2}{2}b$. Alternatively, if the midpoint of its 'slanted' side is height $h$ above its flat side, its area is just $bh$. See if you can think of suitable trapezoids above and below the hyperbola between $x$ and $y$.

1Rock
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Let's denote the area under the hyperbola by $A$, i.e. $A = \ln(x)-\ln(y)$.

Hint1: Here's one way you can go about finding good choices. Let $s(t)$ be the tangent line to $\frac{1}{t}$ at some point $c$. Keeping in mind that $s_0$ in your equation is just $1/t_0$, it's easy to see that

$$s(t) = \frac{2c-t}{c^2}$$

Integrating this from $t=x$ to $t=y$ gives the area of the trapezoid you described. Let's call that area $T_{tan}$. You should find that

$$T_{tan} = \frac{x-y}{2c^2}(4c-x-y)$$

By convexity and playing around with the inequality $T_{tan}<A$, you should find

$$\frac{x-y}{\ln(x)-\ln(y)} < \frac{x-y}{T_{tan}} = \frac{2c^2}{4c-x-y} $$

Can you find a value of $c$ in terms of $x$ and $y$ that gives you the upper bound?

Hint2: Since $\frac{1}{t}$ is decreasing monotonically, you can upper bound its integral over $[x,y]$ using a trapezoidal approximation.

$$ A < \frac{c-y}{2}\cdot\frac{y+c}{yc} + \frac{x-c}{2}\cdot\frac{x+c}{xc} = \frac{c^2x-y^2x+yx^2-c^2y}{2xyc}$$

which, after some factoring, yields

$$\frac{2xyc}{c^2+xy} < \frac{x-y}{\ln(x)-\ln(y)}$$

Again, it remains to pick a good value of $c$.

Philip Hoskins
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