# 2.8. Some Important Inequalities#

Theorem 2.59

Let $$0 < \lambda < 1$$. Then

(2.3)#$t^{\lambda} \leq 1 - \lambda + \lambda t \Forall t \geq 0.$

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

Proof. Define $$f: \RR \to \RR$$ with $$\dom f = \RR_+$$ by

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

Then,

$f'(t) = \lambda - \lambda t^{\lambda -1} = \lambda \left (1 - \frac{1}{t^{1-\lambda}} \right).$

In particular, note that $$f'(1) = 0$$ and:

$\begin{split} f'(t) \begin{cases} > 0, & \text{ if } t > 1\\ < 0, & \text{ if } 0 < t < 1. \end{cases} \end{split}$

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

Thus,

$1 - \lambda + \lambda t - t^{\lambda} \geq 0 \iff t^{\lambda} \leq 1 - \lambda + \lambda t \Forall t \geq 0.$

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 $$p,q \in [1,\infty]$$. Then, $$p$$ and $$q$$ are called conjugate exponents if

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

In particular $$1$$ and $$\infty$$ are conjugate exponents in $$\ERL$$.

Example 2.14 (Conjugate exponents)

1. $$p=1$$ and $$q=\infty$$.

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

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

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

5. $$p=\infty$$ and $$q=1$$.

Theorem 2.60 (Characterization of conjugate exponents)

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

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{split} \begin{aligned} & \frac{1}{p} + \frac{1}{q} = 1\\ & \iff \frac{1}{p} = 1 - \frac{1}{q}\\ & \iff \frac{1}{p} = \frac{q - 1}{q}\\ & \iff p = \frac{q}{q-1}. \end{aligned} \end{split}

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

## 2.8.1. Cauchy Inequality#

Theorem 2.61 (Cauchy inequality)

$2 a b \leq a^2 + b^2.$

Proof. Note that:

\begin{split} \begin{aligned} & (a - b)^2 \geq 0\\ & \iff a^2 -2 a b + b^2 \geq 0 \\ & \iff 2 a b \leq a^2 + b^2. \end{aligned} \end{split}

## 2.8.2. Interpolation Inequality for $$e^x$$.#

Theorem 2.62

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

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

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

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

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

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

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{split} \begin{aligned} &\left (a^p b^{-q} \right)^{\frac{1}{p}} \leq 1 - \frac{1}{p} + \frac{1}{p}a^p b^{-q}\\ &\iff ab^{-\frac{q}{p}} \leq \frac{1}{p}a^p b^{-q} + \frac{1}{q} \\ &\iff ab^{-\frac{q}{p}} b^q \leq \frac{1}{p}a^p + \frac{1}{q} b^q \\ &\iff ab \leq \frac{1}{p}a^p + \frac{1}{q} b^q. \end{aligned} \end{split}

We used the fact that:

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

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

$a^p b^{-q} = 1 \iff a^p = b^q.$

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{split} \begin{aligned} ab &= \exp (\ln (ab))\\ &= \exp ( \ln a + \ln b)\\ &=\exp \left (\frac{1}{p} p \ln a + \frac{1}{q} q \ln b \right )\\ &= \exp \left (\frac{1}{p} \ln (a^p) + \frac{1}{q} \ln (b^q) \right )\\ &\leq \frac{1}{p} \exp (\ln (a^p)) + \frac{1}{q} \exp (\ln (b^q))\\ &= \frac{a^p}{p} + \frac{b^q}{q}. \end{aligned} \end{split}

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 $$1 < p < \infty$$. For any integer $$n \geq 1$$, assume that $$a_1, \dots, a_n$$ and $$b_1, \dots, b_n$$ are non-negative.

Then

$\sum_{k=1}^n a_k b_k \leq \left ( \sum_{k=1}^n a_k^p \right )^{\frac{1}{p}} \left ( \sum_{k=1}^n b_k^q \right )^{\frac{1}{q}}.$

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:

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

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

$\sum_{k=1}^n \frac{a_k}{A} \frac{b_k}{B} \leq \frac{1}{p} \sum_{k=1}^n \frac{a_k^p}{A^p} + \frac{1}{q }\sum_{k=1}^n \frac{b_k^q}{B^q} = \frac{1}{p} + \frac{1}{q} = 1.$

Hence,

$\sum_{k=1}^n a_k b_k \leq A B \leq \left ( \sum_{k=1}^n a_k^p \right )^{\frac{1}{p}} \left ( \sum_{k=1}^n b_k^q \right )^{\frac{1}{q}}$

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.

Theorem 2.65 (AM-GM inequality for two numbers)

Let $$a,b \geq 0$$. Then

$\sqrt{a b} \leq \frac{a + b}{2}.$

Proof. We proceed as follows:

$\begin{split} & (a - b)^2 \geq 0\\ & \iff a^2 -2 a b + b^2 \geq 0 \\ & \iff 2 a b \leq a^2 + b^2\\ & \iff 4 a b \leq a^2 + b^2 + 2 ab\\ & \iff 4 a b \leq (a + b)^2 \\ & \iff ab \leq \left (\frac{a + b}{2} \right )^2\\ & \iff \sqrt{ab} \leq \frac{a + b}{2}. \end{split}$

Theorem 2.66 (Unweighted AM-GM inequality)

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

$\left (\prod_{i=1}^n a_i \right )^{\frac{1}{n}} \leq \frac{1}{n} \left ( \sum_{i=1}^n a_i \right ).$

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.

$$(n) \implies (2n)$$

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

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

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

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

Thus,

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

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

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

Combining, we get the desired result:

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

$$(n) \implies (n-1)$$

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

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

Choose:

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

Then

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

Thus, we have:

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

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

$\begin{split} & \left ( \frac{a_1 + \dots + a_{n-1}}{n-1} \right )^n \geq a_1 \dots a_{n-1} \cdot \frac{a_1 + \dots + a_{n-1}}{n-1}\\ & \implies \left ( \frac{a_1 + \dots + a_{n-1}}{n-1} \right )^{n-1} \geq a_1 \dots a_{n-1} \\ & \implies \frac{a_1 + \dots + a_{n-1}}{n-1} \geq \sqrt[n-1]{a_1 \dots a_{n-1}} \end{split}$

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

By Cauchy induction, the proof is complete.