2.8. Some Important Inequalities#

Theorem 2.58

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 2.59 (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 2.60 (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 2.61

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 2.62 (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 2.63 (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.

We start with the simplest one.

Theorem 2.64 (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 2.65 (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 2.64 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 2.64:

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

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.