Tag: Inequalities

Prove an inequality relating sums and integrals

by RoRi

March 4, 2016

Prove that

$\sum_{k=1}^{n-1} f(k) \leq \int_1^n f(x) \, dx \leq \sum_{k=2}^n f(k)$

where $f$ is a nonnegative, increasing function defined for all $x \geq 1$ .

Use this to deduce the inequalities

$e n^n e^{-n} < n! < e n^{n+1} e^{-n}$

by taking $f(x) = \log x$ .

Proof. Since $f(x)$ is increasing we know $f([x]) \leq f(x) \leq f([x+1])$ (where $[x]$ denotes the greatest integer less than or equal to $x$ ) for all $x \geq 1$ . Define step functions $s$ and $g$ by

$s(x) = f([x]), \qquad t(x) = f([x+1]).$

Then $s$ and $t$ are constant on the open subintervals of the partition

$P = \{ 0,1,2, \ldots, n \}.$

So,

$\begin{align*} \int_1^n s(x) \, dx = \sum_{k=1}^{n-1} s_k (x_k - x_{k-1}) = \sum_{k=1}^{n-1} f(k) \\[9pt] \int_1^n t(x) \, dx = \sum_{k=1}^{n-1} t_k (x_k - x_{k-1}) = \sum_{k=1}^{n-1} f(k+1) = \sum_{k=2}^n f(k). \end{align*}$

Since $f$ is integrable we must have

$\int_1^n s(x) \, dx \leq \int_1^n f(x) \, dx \leq \int_1^n t(x) \, dx$

for every pair of step functions $s \leq f \leq t$ . Hence,

$\sum_{k=1}^{n-1} f(k) \leq \int_1^n f(x) \, dx \leq \sum_{k=2}^n f(k). \qquad \blacksquare$

Next, if we take $f(x) = \log x$ (which is nonnegative and increasing on $x \geq 1$ ) we have

$\begin{align*} &&\sum_{k=1}^{n-1} \log k \leq \int_1^n \log x \, dx \leq \sum_{k=2}^n \log k \\[9pt] \implies && \log(n-1)! \leq n \log n - n + 1 \leq \log n! \\[9pt] \implies && (n-1)! \leq e n^n e^{-n} \leq n!. \end{align*}$

From this we then get two inequalities

$n! \leq en^{n+1} e^{-n} \quad \text{and} \quad e n^n e^{-n} \leq n!.$

Therefore,

$en^n e^{-n} \leq n! \leq e n^{n+1} e^{-n}.$

Prove some statements about integrals of bounded monotonic increasing functions

by RoRi

February 29, 2016

Consider a bounded, monotonic, real-valued function $f$ on the interval $[0,1]$ . The define sequences

$s_n = \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{n}{k} \right), \qquad t_n = \frac{1}{n} \sum_{k=1}^n f \left( \frac{k}{n} \right).$

Prove that
$s_n \leq \int_0^1 f(x) \, dx \leq t_n$

and that

$0 \leq \int_0^1 f(x) \,dx - s_n \leq \frac{f(1) - f(0)}{n}.$
Prove that the two sequences $\{ s_n \}$ and $\{ t_n \}$ converge to $\int_0^1 f(x) \, dx$ .
State and prove a generalization of the above to interval $[a,b]$ .

Proof.First, we define two step functions,
$s(x) = f \left( \frac{[nx]}{n} \right), \qquad t(x) = f \left( \frac{[nx+1]}{n} \right)$

where $[x]$ denotes the greatest integer less than or equal to $x$ . Then we define a partition of $[0,1]$ ,

$P = \left\{ 0, \frac{1}{n}, \frac{2}{n}, \ldots, 1 \right\}.$

For any $x_{k_1} \leq x < x_k$ we have

$s(x) = f \left( \frac{k-1}{n} \right), \qquad t(x) = f \left( \frac{k}{n} \right).$

So, $s(x)$ and $t(x)$ are constant on the open subintervals of the partition $P$ .
Since $f(x)$ is monotonically increasing and $s(x) \leq f(x) \leq t(x)$ for all $x$ (by the definition of $s$ and $t$ ) we have

$\begin{align*} &&\int_0^1 s(x) \, dx \leq \int_0^1 f(x) \, dx \leq \int_0^1 t(x) \, dx \\[9pt] \implies && \sum_{k=0}^{n-1} s_k (x_k - x_{k-1} \leq \int_0^1 f(x) \, dx \leq \sum_{k=0}^{n-1} t_k (x_k - x_{k-1}) && (\text{Def step function integral}) \\[9pt] \implies && \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \leq \int_0^1 f(x) \, dx \leq \frac{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k+1}{n} \right) &&(x_k - x_{k-1} = \frac{1}{n} \forall k) \\[9pt] \implies && \fract{1}{n} \sum_{k=0}^{n-1} f \left( \frac{k}{n} \right) \leq \int_0^1 f(x) \, dx \leq \frac{1}{n} \sum_{k=1}^nf \left( \frac{k}{n} \right) \\[9pt] \implies && s_n \leq \int_0^1 f(x) \, dx \leq t_n && \text{for all } n \\[9pt] \implies && 0 \leq \int_0^1 f(x) \, dx - s_n \leq t_n - s_n \\[9pt] \implies && 0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n}. \qquad \blacksquare \end{align*}$
Proof. From part (a) we have
$0 \leq \int_0^1 f(x) \, dx - s_n \leq \frac{f(1) - f(0)}{n} \quad \implies \quad \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - s_n \right)= 0$

since $\lim_{n \to \infty} \frac{f(1)- f(0)}{n} = 0$ . Since $\int_0^1 f(x) \, dx$ does not depend on $n$ we have

$\int_0^1 f(x) \, dx - \lim_{n \to \infty} s_n = 0 \quad \implies \quad \lim_{n \to \infty} s_n = \int_0^1 f(x) \, dx.$

Therefore,

$\begin{align*} && s_n \leq \int_0^1 f(x) \, dx \leq t_n \\[9pt] \implies && s_n - t_n \leq \int_0^1 f(x) \, dx - t_n \leq 0 \\[9pt] \implies && \frac{f(0)-f(1)}{n} \leq \int_0^1 f(x) \, dx - t_n \leq 0 \\[9pt] \implies && \lim_{n \to \infty} \left( \int_0^1 f(x) \, dx - t_n \right) = 0 \\[9pt] \implies && \lim_{n \to \infty} t_n = \int_0^1 f(x) \, dx. \qquad \blacksquare \end{align*}$
Claim: If $f$ is a real-valued function that is monotonic increasing and bounded on the interval $[a,b]$ , then
$\lim_{n \to \infty} s_n = \int_a^b f(x) \, dx, \qquad \text{and} \qquad \lim_{n \to \infty} t_n = \int_a^b f(x) \, dx$

for $s_n$ and $t_n$ defined as follows:

$s_n = \frac{b-a}{n} \cdot sum_{k=0}^{n-1} f \left( a + k \frac{b-a}{n} \right), \qquad t_n = \frac{b-a}{n} \cdot \sum_{k=1}^n f \left( a + k \frac{b-a}{n} \right).$

Proof. Let

$P = \left\{ a, a + \frac{b-a}{n}, a + \frac{2(b-a)}{n}, \ldots, a + \frac{n(b-a)}{n} = b \right\}$

be a partition of the interval $[a,b]$ . Then, define step functions $s(x)$ and $t(x)$ with $s(x) = f(x_{k-1})$ and $t(x) = f(x_k)$ for $x_{k-1} \leq x < x_k$ . By these definitions we have $s(x) \leq f(x) \leq t(x)$ for all $x \in [a,b]$ (since $f$ is monotonic increasing). Since $f$ is bounded and monotonic increasing it is integrable, and

$\begin{align*} && \int_a^b s(x) \, dx \leq \int_a^b f(x) \, dx \leq \int_a^b t(x) \, dx \\[9pt] \implies && \sum_{k=1}^n s_k (x_k - x_{k-1}) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n t_k (x_k - x_{k-1}) \\[9pt] \implies && \sum_{k=1}^n s_k \left( \frac{b-a}{n} \right) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n t_k \left( \frac{b-a}{n} \right). \end{align*}$

And, since $s_k = f(x_{k-1} = f \left( a + \frac{(k-1)(b-a)}{n} \right)$ , and $t_k = f(x_k) = f \left( a + \frac{k(b-a)}{n} \right)$ we have

$\begin{align*} \implies && \sum_{k=1}^n f \left( a + \frac{(k-1)(b-a)}{n} \right) \left( \frac{b-a}{n} \right) \leq \int_a^b f(x) \, dx \leq \sum_{k=1}^n f \left( a + \frac{k(b-a)}{n} \right) \left( \frac{b-a}{n} \right) \\[9pt] \implies && \left( \frac{b-a}{n} \right) \sum_{k=0}^{n-1} f \left( a+ \frac{k(b-a)}{n} \right) \leq \int_a^b f(x) \, dx \leq \left( \frac{b-a}{n} \right) \sum_{k=1}^n f \left( a+ \frac{k(b-a)}{n} \right). \qquad \blacksquare \end{align*}$

Sketch inequalities in the complex plane

by RoRi

February 27, 2016

Sketch each of the following sets of complex numbers $z$ that satisfy the given inequalities:

$|2z + 3| < 1$ .
$|z+1| < |z-1|$ .
$|z-i| \leq |z+i|$ .
$|z| \leq |2z+1|$ .

Letting $z = x+ iy$ we have,
$\begin{align*} |2z+3| < 1 && \implies && |2x+3 + 2yi| &< 1 \\ && \implies && \sqrt{(2x+3)^2 + 4y^2} &< 1 \\ && \implies && (2x+3)^2 + 4y^2 < 1. \end{align*}$

This is a disk of radius $\frac{1}{2}$ centered at $\left(-\frac{3}{2}, 0 \right)$ . The sketch is as follows:
Letting $z = x+ iy$ we have,
$\begin{align*} |z+1| < |z-1| && \implies && \sqrt{(x+1)^2 +y^2} &< \sqrt{(x-1)^2 + y^2} \\ && \implies && x^2 + 2x + 1 + y^2 &< x^2 -2x + 1 + y^2 \\ && \implies && 4x &< 0 \\ && \implies && x &< 0. \end{align*}$

This is the half-plane with negative real part. The sketch is as follows:
Letting $z = x + iy$ we have,
$\begin{align*} |z-i| \leq |z+i| && \implies && \sqrt{x^2 + (y-1)^2} &\leq \sqrt{x^2 + (y+1)^2} \\ && \implies && x^2 + y^2 - 2y + 1 &\leq x^2 + y^2 + 2y + 1 \\ && \implies && y & \geq 0. \end{align*}$

This is the half-plane with positive imaginary part. The sketch is as follows:
Letting $z = x+ iy$ we have,
$\begin{align*} |z| \leq |2z+1| && \implies && \sqrt{x^2+y^2} &\leq \sqrt{(2x+1)^2 + (2y)^2} \\ && \implies && x^2 + y^2 &\leq 4x^2 + 4x + 1 + 4y^2 \\ && \implies && 0 &\leq 3x^2 + 4x + 1 + 3y^2 \\ && \implies && 0 &\leq x^2 + \frac{4}{3}x + \frac{4}{9} + y^2 - \frac{1}{9} \\ && \implies && \frac{1}{9} &\leq \left( x +\frac{2}{3} \right)^2 + y^2. \end{align*}$

This is the region outside the disk of radius $\frac{1}{3}$ centered at the point $\left( -\frac{2}{3}, 0 \right)$ . The sketch is as follows:

Prove an inequality for the integral of 1 / (1 + x⁴)

by RoRi

December 21, 2015

Prove that

$0.493948 < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < 0.493958.$

(Note: I cannot get the bounds Apostol asks for. I prove a different set below. I cannot figure out if it is a mistake in the book or not.)

Proof. Using the algebraic identity, valid for $0 < x < 1$ ,

$\frac{1}{1-x} = 1+x+x^2+\cdots + x^n + \frac{x^{n+1}}{1-x},$

we obtain

$\frac{1}{1+x^4} = 1 - x^4 + x^8 - x^{12} + \cdots + (-1)^n x^{4n} + (-1)^{n+1} \frac{x^{4n+4}}{1+x^4}.$

Therefore, integrating term by term,

$\begin{align*} \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx &= \int_0^{\frac{1}{2}} \left( 1 - x^4 + x^8 + \cdots + (-1)^n x^{4n} + (-1)^{n+1} \frac{x^{4n+4}}{1+x^4} \right) \, dx \\[9pt] &= \left( x - \frac{x^5}{5} + \frac{x^9}{9} - \cdots + (-1)^n \frac{x^{4n+1}}{4n+1} \right)\Bigr \rvert_0^{\frac{1}{2}} + (-1)^{n+1} \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx. \end{align*}$

Furthermore, we have

$\int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx < \int_0^{\frac{1}{2}} x^{4n+4} \, dx = \left( \frac{1}{2} \right)^{4n+5} \left( \frac{1}{4n+5} \right).$

Taking $n = 2$ , we then have

$\int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx = \left( \frac{1}{2} \right) - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} - \int_0^{\frac{1}{2}} \frac{x^{4n+4}}{1+x^4} \, dx.$

From the inequality for this integral we then have

$\begin{align*} &\frac{1}{2} - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} - \left( \frac{1}{2} \right)^{4n+5} \left( \frac{1}{4n+5} \right) < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx\\ \intertext{and} &\int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < \frac{1}{2} - \frac{(1/2)^5}{5} + \frac{(1/2)^9}{9} \\[9pt] \implies & 0.493958 < \int_0^{\frac{1}{2}} \frac{1}{1+x^4} \, dx < 0.49367. \qquad \blacksquare \end{align*}$

Prove an expression for the integral from 0 to 1 of (1 + x³⁰) / (1+x⁶⁰)

by RoRi

December 21, 2015

Prove that there exists a number $c$ with $0 < c < 1$ such that

$\int_0^1 \frac{1+x^{30}}{1+x^{60}} \, dx = 1 + \frac{c}{31}.$

Proof. For $0 < x < 1$ we have

$1 < \frac{1+x^{30}}{1+x^{60}} < 1 + x^{30}.$

(Since for $0 < x < 1$ we know $1+x^{60} < 1+x^{30}$ .) Therefore, integrating the terms in the inequality from 0 to 1,

$\begin{align*} &\int_0^1 \, dx < \int_0^1 \frac{1+x^{30}}{1+x^{60}} \, dx < \int_0^1 (1+x^{30}) \, dt \\[9pt] \implies & 1 < \int_0^1 \frac{1+x^{30}}{1+x^{60}} \, dx < 1 + \frac{1}{31} \\[9pt] \implies & \int_0^1 \frac{1+x^{30}}{1+x^{60}} \, dx = 1 + \frac{c}{31} \qquad \text{for some } 0 < c < 1. \qquad \blacksquare \end{align*}$

Use Taylor polynomials to approximate the nonzero root of arctan x = x²

by RoRi

December 21, 2015

Show that $r = \frac{\sqrt{21}-3}{2}$ is an approximation of the nonzero root of the equation
$\arctan x = x^2$

using the cubic Taylor polynomial approximation to $\arctan x$ .
Given that
$\sqrt{21} < 4.6 \qquad \text{and} \qquad 2^{16} = 65536$

prove that the number $r$ from part (a) satisfies

$|r^2 - \arctan x| < \frac{7}{100}.$

Determine if $(r^2 - \arctan r)$ is positive or negative and prove the result.

Proof. From a previous exercise (Section 7.8, Exercise #3) we know
$\arctan x = x - \frac{x^3}{3} + E_{2n}(x).$

So, to approximate the nonzero root of $x^2 - \arctan x$ we have

$\begin{align*} x^2 - x + \frac{x^3}{3} \approx 0 && \implies && x^2 + 3x - 3 &\approx 0 \\ && \implies && x &\approx \frac{\sqrt{21}-3}{2}. \qquad \blacksquare \end{align*}$
We know from the same previous exercise we used in part (a) that the error term $E_{2n}(x)$ for $\arctan x$ satisfies the inequality
$|E_{2n}(x)| \leq \frac{x^{2n+1}}{2n+1}.$

Using the values for $\sqrt{21}$ and $2^{16}$ given we have

$\begin{align*} \left|E_5 \left( \frac{\sqrt{21}-3}{2} \right)\right| &\leq \left| \frac{\left( \frac{\sqrt{21}-3}{2} \right)^5}{5} \right| \leq \left| \frac{0.8^5}{5} \right| = \frac{(4/5)^5}{5} \\[9pt] &= \frac{2^{10}}{5^6} = \frac{2^{16}}{10^6} = \frac{65536}{1000000} < \frac{7}{100}. \qquad \blaacksquare \end{align*}$

Use Taylor polynomials to approximate the nonzero root of x²=sin x

by RoRi

December 21, 2015

Using the cubic Taylor polynomial approximation of $\sin x$ , show that the nonzero root of the equation
$x^2 = \sin x$

is approximated by $r = \sqrt{15} - 3$ .
Using part (a) show that
$| \sin r - r^2 | < \frac{1}{200},$

given that $\sqrt{15} - 3 < 0.9$ . Determine whether $(\sin r - r^2)$ is positive or negative and prove the result.

Proof. The cubic Taylor polynomial approximation of $\sin x$ is
$\sin x = x - \frac{x^3}{3!} + E_{2n} (x).$

This implies

$x^2 - \sin x \approx x^2 - x + \frac{x^3}{6}.$

Therefore, we can approximate the nonzero root by

$\begin{align*} x^2 - \sin x = 0 && \implies && x^2 - x + \frac{x^3}{6} &\approx 0 \\ && \implies && 6x^2 + x - 1 &\approx 0 \\ && \implies && x &\approx \sqrt{15} - 3. \qquad \blacksquare \end{align*}$
Proof. We know from this exercise (Section 7.8, Exercise #1) that for $\sin x$ we have
$|E_{2n} (x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}.$

So, for $n = 2$ , and using the given inequality $\sqrt{15} - 3 < 0.9$ , we have

$|E_{2n} (x)| \leq \frac{|0.9|^5}{120} = \frac{9^5}{10^5 (120)} < \frac{1}{200}.$

Furthermore, $(\sin r - r^2) > 0$ since

$\sin r - r^2 = E_{2n}(x) = \frac{x^5}{5!} - \frac{x^7}{7!}+ \cdots$

with the absolute value of each term in the sum strictly less than the absolute value of the previous term (since $x < 1$ and $(n+2)! > n!$ ). Thus, each pair is positive, so the whole series is positive $. \qquad \blacksquare$

Prove an inequality for the error of the Taylor polynomial of arctan x

by RoRi

December 21, 2015

Prove that the error term in the Taylor expansion of $\arctan x$ satisfies the following inequality.

$\arctan(x) = \sum_{k=0}^{n-1} \frac{(-1)^k x^{2k+1}}{2k+1} + E_{2n} (x), \qquad |E_{2n}(x)| \leq \frac{x^{2n+1}}{2n+1}, \quad \text{if } 0 \leq x \leq 1.$

Proof. To prove this we will work directly from the definition of the error as an integral,

$E_n (x) = \frac{1}{n!} \int_a^x (x-t)^n f^{(n+1)}(t) \, dt.$

We know for $0 \leq x \leq 1$ we have, (we need $0 \leq x \leq 1$ for the expansion of $\frac{1}{1+x^2}$ to be valid),

$\begin{align*} \arctan x &= \int_0^x \frac{1}{1+t^2} \, dt \\[9pt] &= \int_0^x \left( 1 - t^2 + t^4 - t^6 + \cdots + (-1)^{n-1} t^{2n-2} + \frac{(-1)^n t^{2n}}{1+t^2} \right) \, dt \\[9pt] &= \int_0^x \left( 1 - t^2 + t^4 - t^6 + \cdots + (-1)^{n-1} t^{2n-2} \right) \, dt + \int_0^x \frac{(-1)^n t^{2n}}{1+t^2} \, dt \\[9pt] &= \sum_{k=0}^{n-1} \frac{(-1)^k x^{2k+1}}{2k+1} + \int_0^x \frac{(-1)^n t^{2n}}{1+t^2} \, dt \\[9pt] &= \sum_{k=0}^{n-1} \frac{(-1)^k x^{2k+1}}{2k+1} + E_{2n}(x). \end{align*}$

Therefore we have

$E_{2n}(x) = \int_0^x \frac{(-1)^n t^{2n}}{1+t^2}.$

So, we can bound the error term by bounding the integral,

$\begin{align*} | E_{2n} (x) | &= \left| \int_0^x \frac{(-1)^n t^{2n}}{1+t^2} \, dt \right| \\[9pt] &\leq \int_0^x \frac{t^{2n}}{1+t^2} \, dt \\[9pt] &\leq \int_0^x t^{2n} \, dt &(1+t^2 \geq 1) \\[9pt] &= \frac{x^{2n+1}}{2n+1}. \qquad \blacksquare \end{align*}$

Prove an inequality for the error of the Taylor polynomial of sin x

by RoRi

December 21, 2015

Prove that the error of the Taylor expansion of $\sin x$ satisfies the following inequality.

$\sin x = \sum_{k=1}^n \frac{(-1)^{k-1} x^{2k-1}}{(2k-1)!} + E_{2n}(x), \qquad |E_{2n}(x)| \leq \frac{|x|^{2n+1}}{(2n+1)!}.$

Proof. Since the derivatives of $\sin x$ are always $\cos x, -\sin x, -\cos x$ , or $\sin x$ we know that for $f(x) = \sin x$ we have $|f^{(n+1)}(x)| \leq 1$ . (In other words, the $n+1$ st derivative is bounded above by 1 and below by $-1$ .) Therefore, we can apply Theorem 7.7 (p. 280 of Apostol) to estimate the error in Taylor’s formula at $a= 0$ with $m = -1$ and $M = 1$ . For $x > 0$ this gives us

$\begin{alignat*}{3} & m \frac{(x-a)^{n+1}}{(n+1)!} &\leq \ \ E_n (x) &\leq M \frac{(x-a)^{n+1}}{(n+1)!} \\[9pt] \implies & -\frac{x^{n+1}}{(n+1)!} &\leq \ \ E_n (x) &\leq \frac{x^{n+1}}{(n+1)!} \\[9pt] \implies & -\frac{x^{2n+1}}{(2n+1)!} &\leq \ \ E_{2n} (x) &\leq \frac{x^{2n+1}}{(2n+1)!} \\[9pt] \implies & \phantom{-}|E_{2n}(x)| &\leq \frac{x^{2n+1}}{(2n+1)!}. \end{alignat*}$

Next, (from the second part of Theorem 7.7) if $x < 0$ we have

$\begin{alignat*}{3} & m \frac{(a-x)^{n+1}}{(n+1)!} &\leq (-1)^{n+1} E_n (x) &\leq M \frac{(a-x)^{n+1}}{(n+1)!} \\[9pt] \implies & - \frac{(-x)^{n+1}}{(n+1)!} &\leq (-1)^{n+1} E_n (x) &\leq \frac{(-x)^{n+1}}{(n+1)!} \\[9pt] \implies & - \frac{(-x)^{2n+1}}{(2n+1)!} &\leq (-1)^{2n+1} E_{2n}(x) &\leq \frac{(-x)^{2n+1}}{(2n+1)!} \\[9pt] \implies & - \frac{(-x)^{2n+1}}{(2n+1)!} &\leq -E_{2n} (x) &\leq \frac{(-x)^{2n+1}}{(2n+1)!} \\[9pt] \implies & \phantom{-} |E_{2n} (x)| &\leq \frac{x^{2n+1}}{(2n+1)!}. \qquad \blacksquare \end{alignat*}$

Prove an inequality of exponentials

by RoRi

December 20, 2015

For all $x, y > 0$ and for any constants $a,b$ such that $0 < a < b$ prove that

$\big( x^b + y^b \big)^{\frac{1}{b}} < \big( x^a + y^a \big)^{\frac{1}{a}}.$

Proof. We want to consider the function

$f(t) = (x^t + y^t)^{\frac{1}{t}}.$

If we can show this function is decreasing on the positive real axis then we establish the inequality since this would mean that if $0 < a < b$ then

$f(b) < f(a) \quad \implies \quad (x^b+y^b)^{\frac{1}{b}} < (x^a+y^a)^{\frac{1}{a}}.$

(So, the trick here is to think of this as a function of the exponent. The $x$ and $y$ are some positive fixed constants.) To take the derivative of $f(t)$ we use logarithmic differentiation,

$\begin{align*} &&f(t) &= (x^t+y^t)^{\frac{1}{t}} \\[10pt] \implies &&\log f(t) &= \frac{1}{t} \log (x^t+y^t) \\[10pt] \implies &&\frac{d}{dt} (\log f(t)) &= \frac{d}{dt} \left( \frac{1}{t} \log (x^t+y^t) \right) \\[10pt] \implies &&\frac{f'(t)}{f(t)} &= -\frac{1}{t^2} \log (x^t+y^t) + \frac{1}{t} \left( \frac{1}{x^t+y^t} \right) \left( x^t \log x + y^t \log y \right) \\[10pt] \implies &&\frac{f'(t)}{f(t)} &= \frac{-\log(x^t+y^t)}{t^2} + \frac{x^t \log x + y^t \log y}{t (x^t+y^t)}. \end{align*}$

Multiplying both sides by $f(t)$ we then obtain

$\begin{align*} f'(t) &= \left( \frac{-\log(x^t+y^t)}{t^2} + \frac{x^t \log x + y^t \log y}{t(x^t+y^t)} \right) (x^t+y^t)^{\frac{1}{t}} \\[10pt] &= (x^t + y^t)^{\frac{1}{t}} \left( \frac{t(x^t \log x + y^t \log y) - (x^t+y^t)\log(x^t+y^t)}{t^2 (x^t+y^t)} \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t} - 1}}{t^2} \right) \left( x^t t \log x + y^t t \log y - (x^t + y^t)\log(x^t+y^t) \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t}-1}}{t^2} \right) \left( x^t \log x^t + y^t \log y^t - x^t \log (x^t+y^t) - y^t \log (x^t+y^t) \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t} - 1}}{t^2} \right) \left( x^t \log \left(\frac{x^t}{x^t+y^t}\right) + y^t \log \left( \frac{y^t}{x^t+y^t} \right) \right). \end{align*}$

Now we can conclude that $f'(t) < 0$ for all $t > 0$ since the first term in the product

$\frac{(x^t+y^t)^{\frac{1}{t}-1}}{t^2} > 0.$

Since $x, y > 0$ (any real power of a positive number is still positive) and $t^2 > 0$ . For the second term we have

$x^t \log \left( \frac{x^t}{x^t+y^t} \right) + y^t \log \left( \frac{y^t}{x^t+y^t} \right) < 0$

since $x^t$ and $y^t$ are positive, but both logarithms are negative. We know these logarithms are negative since

$\frac{x^t}{x^t+y^t} < 1 \quad \text{and} \quad \frac{y^t}{x^t+y^t} < 1$

implies

$\log \left( \frac{x^t}{x^t+y^t} \right) < 0 \quad \text{and} \quad \log \left( \frac{y^t}{x^t+y^t} \right) < 0.$

Hence, $f'(t) < 0$ for all $t > 0$ . This means $f(t)$ is a decreasing function. Therefore, if $0 < a < b$ then we have

$f(b) < f(a) \quad \implies \quad \big( x^b + y^b \big)^{\frac{1}{b}} < \big( x^a + y^a \big)^{\frac{1}{a}}. \qquad \blacksquare$