2.8. Some Important Inequalities

Theorem 2.59

Let $0<\lambda <1$. Then

(2.3)$$t^{\lambda }\le 1-\lambda +\lambda t\mathrm{\forall }t\ge 0.$$

The inequality becomes equality only when $t=1$.

Proof. Define $f:\mathbb{R}\to \mathbb{R}$ with $\mathrm{dom}f={\mathbb{R}}_{+}$ by

$$f(t)\triangleq 1-\lambda +\lambda t-t^{\lambda }.$$

Then,

$$f^{\prime }(t)=\lambda -\lambda t^{\lambda -1}=\lambda (1-\frac{1}{t^{1-\lambda }}).$$

In particular, note that $f^{\prime }(1)=0$ and:

$$\begin{array}{r}{f}^{\prime }(t)\{\begin{array}{ll}>0,& \text{ if }t>1\\ <0,& \text{ if }0<t<1.\end{array}\end{array}$$

Thus, $f$ has a minimum value at $t=1$ which is $f(1)=0$. Thus, $f(t)\ge 0$ for all $t\ge 0$.

Thus,

$$1-\lambda +\lambda t-t^{\lambda }\ge 0⟺t^{\lambda }\le 1-\lambda +\lambda t\mathrm{\forall }t\ge 0.$$

For $t=1$,

$$1-\lambda +\lambda t=1=1^{\lambda }.$$

Thus, the inequality indeed reduces to equality at $t=1$.

Definition 2.89 (Conjugate exponent)

Let $p,q$ be extended real numbers with $p,q\in [1,\mathrm{\infty }]$. Then, $p$ and $q$ are called conjugate exponents if

$$\frac{1}{p}+\frac{1}{q}=1.$$

In particular $1$ and $\mathrm{\infty }$ are conjugate exponents in $\bar{\mathbb{R}}$.

Example 2.14 (Conjugate exponents)

1. $p=1$ and $q=\mathrm{\infty }$.

2. $p=2/3$ and $q=3$.

3. $p=2$ and $q=2$.

4. $p=3$ and $q=2/3$.

5. $p=\mathrm{\infty }$ and $q=1$.

Theorem 2.60 (Characterization of conjugate exponents)

Let $p,q$ be extended real numbers with $p,q\in [1,\mathrm{\infty }]$.

The following are equivalent:

1. $p$ and $q$ are conjugate exponents.

2. $\frac{1}{p}=\frac{q-1}{q}$.

3. $p=\frac{q}{q-1}$.

4. $\frac{1}{q}=\frac{p-1}{p}$.

5. $q=\frac{p}{p-1}$.

Proof. We get these results by simple arithmetic.

$$\begin{array}{r}\begin{array}{rl}& \frac{1}{p}+\frac{1}{q}=1\\ & ⟺\frac{1}{p}=1-\frac{1}{q}\\ & ⟺\frac{1}{p}=\frac{q-1}{q}\\ & ⟺p=\frac{q}{q-1}.\end{array}\end{array}$$

The other two equalities are obtained by simply interchanging $p$ with $q$.

2.8.1. Cauchy Inequality

Theorem 2.61 (Cauchy inequality)

$$2ab\le a^2+b^2.$$

Proof. Note that:

$$\begin{array}{r}\begin{array}{rl}& (a-b{)}^2\ge 0\\ & ⟺a^2-2ab+b^2\ge 0\\ & ⟺2ab\le a^2+b^2.\end{array}\end{array}$$

2.8.2. Interpolation Inequality for $e^x$.

Theorem 2.62

If $t\in [0,1]$ then:

$$e^{ta+(1-t)b}\le t{e}^a+(1-t)e^b.$$

Proof. This is a direct implication of the fact that $e^x$ is convex.

2.8.3. Young’s Inequality

Theorem 2.63 (Young’s inequality)

Let $p\in (1,\mathrm{\infty })$. Let $q$ be the conjugate exponent of $p$. Let $a,b\ge 0$, the following holds true:

$$ab\le \frac{a^p}{p}+\frac{b^q}{q}.$$

Note that for the special case of $p=q=2$, we obtain

$$ab\le \frac{a^2}{2}+\frac{b^2}{2}$$

which is same as the Cauchy inequality. Thus, Young’s inequality is a generalization of Cauchy inequality.

Proof. For $a=0$ or $b=0$, the inequality is obvious. We shall now consider the case where $a>0$ and $b>0$.

Let $\lambda =\frac{1}{p}$. Let $t=a^p{b}^{-q}$. Putting this in (2.3), we obtain:

$$\begin{array}{r}\begin{array}{rl}& {\left(a^p{b}^{-q}\right)}^{\frac{1}{p}}\le 1-\frac{1}{p}+\frac{1}{p}{a}^p{b}^{-q}\\ & ⟺a{b}^{-\frac{q}{p}}\le \frac{1}{p}{a}^p{b}^{-q}+\frac{1}{q}\\ & ⟺a{b}^{-\frac{q}{p}}{b}^q\le \frac{1}{p}{a}^p+\frac{1}{q}{b}^q\\ & ⟺ab\le \frac{1}{p}{a}^p+\frac{1}{q}{b}^q.\end{array}\end{array}$$

We used the fact that:

$$-\frac{q}{p}+q=q(-\frac{1}{p}+1)=q\frac{1}{q}=1.$$

Recall that (2.3) is an equality only if $t=1$ giving us:

$$a^p{b}^{-q}=1⟺a^p=b^q.$$

Following is an alternative proof. This proof exploits the fact that $e^x$ is convex.

Proof. For $a=0$ or $b=0$, the inequality is obvious. We shall now consider the case where $a>0$ and $b>0$.

$$\begin{array}{r}\begin{array}{rl}ab& =\exp (\ln (ab))\\ & =\exp (\ln a+\ln b)\\ & =\exp (\frac{1}{p}p\ln a+\frac{1}{q}q\ln b)\\ & =\exp (\frac{1}{p}\ln (a^p)+\frac{1}{q}\ln (b^q))\\ & \le \frac{1}{p}\exp (\ln (a^p))+\frac{1}{q}\exp (\ln (b^q))\\ & =\frac{a^p}{p}+\frac{b^q}{q}.\end{array}\end{array}$$

In this derivation, we used the fact that $e^x$ is strictly convex.

2.8.4. Hölder’s Inequality

Theorem 2.64 (Hölder’s inequality)

Let $p,q$ be conjugate exponents with $1<p<\mathrm{\infty }$. For any integer $n\ge 1$, assume that $a_1,\dots ,a_n$ and $b_1,\dots ,b_n$ are non-negative.

Then

$$\sum _{k=1}^n{a}_k{b}_k\le {\left(\sum _{k=1}^n{a}_k^p\right)}^{\frac{1}{p}}{\left(\sum _{k=1}^n{b}_k^q\right)}^{\frac{1}{q}}.$$

Proof. Let

$$A={\left(\sum _{k=1}^n{a}_k^p\right)}^{\frac{1}{p}}\text{ and }B={\left(\sum _{k=1}^n{b}_k^q\right)}^{\frac{1}{q}}.$$

If $AB=0$ (i.e., either $A=0$ or $B=0$ or both) then either all $a_i$ or all $b_i$ must be zero and the inequality is obvious.

Now, consider the case $AB>0$; i.e., $A>0$ and $B>0$.

Observe that:

$$\sum _{k=1}^n\frac{a_k^p}{A^p}=1=\sum _{k=1}^n\frac{b_k^q}{B^q}.$$

Letting h $a=\frac{a_k}{A}$ and $b=\frac{b_k}{B}$ and applying Young's inequality, we get:

$$\frac{a_k}{A}\frac{b_k}{B}\le \frac{a_k^p}{p{A}^p}+\frac{b_k^q}{q{B}^q}.$$

Summing over $1\le k\le n$, we obtain:

$$\sum _{k=1}^n\frac{a_k}{A}\frac{b_k}{B}\le \frac{1}{p}\sum _{k=1}^n\frac{a_k^p}{A^p}+\frac{1}{q}\sum _{k=1}^n\frac{b_k^q}{B^q}=\frac{1}{p}+\frac{1}{q}=1.$$

Hence,

$$\sum _{k=1}^n{a}_k{b}_k\le AB\le {\left(\sum _{k=1}^n{a}_k^p\right)}^{\frac{1}{p}}{\left(\sum _{k=1}^n{b}_k^q\right)}^{\frac{1}{q}}$$

as desired.

2.8.5. AM-GM Inequalities

These inequalities establish relationship between arithmetic mean and geometric mean of a set of nonnegative real numbers.

We start with the simplest one.

Theorem 2.65 (AM-GM inequality for two numbers)

Let $a,b\ge 0$. Then

$$\sqrt{ab}\le \frac{a+b}{2}.$$

Proof. We proceed as follows:

$$\begin{array}{rl}& (a-b{)}^2\ge 0\\ & ⟺a^2-2ab+b^2\ge 0\\ & ⟺2ab\le a^2+b^2\\ & ⟺4ab\le a^2+b^2+2ab\\ & ⟺4ab\le (a+b{)}^2\\ & ⟺ab\le {\left(\frac{a+b}{2}\right)}^2\\ & ⟺\sqrt{ab}\le \frac{a+b}{2}.\end{array}$$

Theorem 2.66 (Unweighted AM-GM inequality)

Let $a_1,a_2,\dots ,a_n\ge 0$. Then

$${\left(\prod _{i=1}^n{a}_i\right)}^{\frac{1}{n}}\le \frac{1}{n}\left(\sum _{i=1}^n{a}_i\right).$$

Proof. If any of the numbers is 0, then the geometric mean is 0 and the inequality is satisfied trivially. Thus, we shall assume that all numbers are positive.

We prove this by Cauchy induction.

1. The base case for $n=2$ is proved in Theorem 2.65 above.

2. We show that if the inequality is true for some $n$, then it is also true for $2n$.

3. We then show that if the inequality is true for some $n$, then it is true for $n-1$ too.

4. Then, by principle of Cauchy induction, the proof is complete.

$(n)⟹(2n)$

Assume that the inequality holds true for any set of $n$ positive numbers. Now for $2n$ numbers:

$$\frac{a_1+\cdots +a_{2n}}{2n}=\frac{\frac{a_1+\cdots +a_n}{n}+\frac{a_{n+1}+\cdots +a_{2n}}{n}}{2}.$$

Since the inequality holds true for $n$, hence:

$$\frac{a_1+\cdots +a_n}{n}\ge \sqrt[n]{a_1\dots a_n}\text{ and }\frac{a_{n+1}+\cdots +a_{2n}}{n}\ge \sqrt[n]{a_{n+1}\dots a_{2n}}.$$

Thus,

$$\frac{a_1+\cdots +a_{2n}}{2n}\ge \frac{\sqrt[n]{a_1\dots a_n}+\sqrt[n]{a_{n+1}\dots a_{2n}}}{2}.$$

The R.H.S. is an arithmetic mean of 2 numbers. Applying Theorem 2.65:

$$\frac{\sqrt[n]{a_1\dots a_n}+\sqrt[n]{a_{n+1}\dots a_{2n}}}{2}\ge \sqrt{\sqrt[n]{a_1\dots a_n}\sqrt[n]{a_{n+1}\dots a_{2n}}}=\sqrt[2n]{a_1\dots a_{2n}}.$$

Combining, we get the desired result:

$$\frac{a_1+\cdots +a_{2n}}{2n}\ge \sqrt[2n]{a_1\dots a_{2n}}.$$

$(n)⟹(n-1)$

By induction hypothesis, for any $n$ positive numbers, we have:

$$\frac{a_1+\cdots +a_n}{n}\ge \sqrt[n]{a_1\dots a_n}.$$

Choose:

$$a_n=\frac{a_1+\cdots +a_{n-1}}{n-1}.$$

Then

$$\frac{a_1+\cdots +a_n}{n}=\frac{a_1+\cdots +\frac{a_1+\cdots +a_{n-1}}{n-1}}{n}=\frac{a_1+\cdots +a_{n-1}}{n-1}.$$

Thus, we have:

$$\frac{a_1+\cdots +a_{n-1}}{n-1}\ge \sqrt[n]{a_1\dots a_{n-1}\cdot \frac{a_1+\cdots +a_{n-1}}{n-1}}.$$

Taking $n$-th power on both sides, we get:

$$\begin{array}{rl}& {\left(\frac{a_1+\cdots +a_{n-1}}{n-1}\right)}^n\ge a_1\dots a_{n-1}\cdot \frac{a_1+\cdots +a_{n-1}}{n-1}\\ & ⟹{\left(\frac{a_1+\cdots +a_{n-1}}{n-1}\right)}^{n-1}\ge a_1\dots a_{n-1}\\ & ⟹\frac{a_1+\cdots +a_{n-1}}{n-1}\ge \sqrt[n-1]{a_1\dots a_{n-1}}\end{array}$$

as desired. The division by $\frac{a_1+\cdots +a_{n-1}}{n-1}$ is valid since it is a positive number.

By Cauchy induction, the proof is complete.