Classical Inequalities

The classical inequalities are a number of generalized inequalities that have wide use in algebra. They are often used for determining minimum and maximum values of functions. Although maximums and minimums can be found using methods from calculus, the application of a classical inequality is often a simpler approach.

Main Article: AM-GM Inequality

The AM-GM inequality relates the arithmetic mean (AM) to the geometric mean (GM). The AM-GM inequality is commonly used in competition math to find the maximum or minimum value of a multi-variable function or expression.

Given non-negative real numbers $a_1,a_2,\cdots,a_n,$ the geometric mean of these numbers cannot exceed the arithmetic mean, and they will be equal if and only if all the chosen numbers are equal. That is,

$$\frac{ a_1 + a_2 + \cdots + a_n } { n}\ge\sqrt[n] { a_1 a_2 \ldots a_n}$$

with equality if and only if $a_1=a_2=\cdots =a_n$.

More precisely,

$$\frac{\displaystyle \sum _{ i=1 }^{ n }{ { a }_{ i } }}{n} \ge \sqrt [ n ]{ \prod _{ i=1 }^{ n }{ { a }_{ i } } } . \ _\square$$

If $x,$ $y,$ and $z$ are positive real numbers such that $xyz=1,$ then what is the minimum value of $4xy^2+2x^2y+27z^3 ?$

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Notice that the given equation contains a product of variables, and the expression to minimize is a sum of terms. These expressions can be related through the AM-GM inequality:

$$\begin{aligned} \frac{4xy^2+2x^2y+27z^3}{3} &\ge \sqrt[3]{\left(4xy^2\right) \left(2x^2y\right) \left(27z^3\right)} \\ \\ \frac{4xy^2+2x^2y+27z^3}{3} &\ge \sqrt[3]{216x^3y^3z^3} \\ \\ \frac{4xy^2+2x^2y+27z^3}{3} &\ge 6xyz \\ \\ 4xy^2+2x^2y+27z^3 &\ge 18. \end{aligned}$$

Thus, the minimum value of the expression is $\boxed{18}.$

Furthermore, this minimum is achieved only if

$$4xy^2=2x^2y=27z^3.$$

Solving this system of equations gives $x=\sqrt[3]{6},$ $y=\frac{\sqrt[3]{6}}{2},$ and $z=\frac{\sqrt[3]{6}}{3}.$ $_\square$

The weighted AM-GM inequality is a generalization of the AM-GM inequality. It relates a weighted arithmetic mean to a weighted geometric mean.

For non-negative numbers $a_1,...,a_n$ and non-negative weights $\omega_1,...,\omega_n,$

$$\large\dfrac {\displaystyle\sum \omega_i a_i } { \displaystyle\sum \omega_i } \geq \sqrt[\sum\omega_i]{\displaystyle\prod a_i ^ {\omega_i }}$$

with equality if and only if $a_1=a_2=\cdots =a_n.$

Given that $x,$ $y,$ and $z$ are positive real numbers such that $x^3y^2z=108,$ what is the minimum value of $x+y+z?$

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Notice that the equation contains a product of powers, and the expression to minimize contains a sum of linear terms. The strategy is to assign coefficients to each variable so that these expressions can be related via the weighted AM-GM inequality:

$$\begin{aligned} \left(\frac{1}{3}x\right)^3 \left(\frac{1}{2}y\right)^2 z &= 108 \left(\frac{1}{3}\right)^3 \left(\frac{1}{2}\right)^2 \\ &= 1. \end{aligned}$$

Now these expressions can be related with the weighted AM-GM inequality:

$$\begin{aligned} \frac{ 3\left(\frac{1}{3}x\right) + 2\left(\frac{1}{2}x\right) + 1z }{3+2+1} &\ge \sqrt[3+2+1]{ \left(\frac{1}{3}x\right)^3 \left(\frac{1}{2}y\right)^2 z } \\ \\ \frac{x+y+z}{6} &\ge \sqrt[6]{1} \\ \\ x+y+z &\ge 6. \end{aligned}$$

Thus, the minimum value of $x+y+z$ is $\boxed{6}.$ Furthermore, this minimum is achieved only if

$$\frac{1}{3}x=\frac{1}{2}y=z.$$

Solving this system of equations gives $x=3,$ $y=2,$ and $z=1.$ $_\square$

Main Article: Cauchy-Schwarz inequality

The Cauchy-Schwarz inequality relates a product of sums of squares to the square of a sum of products. Like the AM-GM inequality, the Cauchy-Schwarz inequality is commonly used in competition math to find the minimum or maximum value of a multi-variable function or expression.

Given real numbers $a_1,a_2,\cdots,a_n$ and $b_1,b_2,\cdots,b_n$, we have

$$\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)\ge \left(\sum_{i=1}^n a_ib_i\right)^2,$$

where equality holds if and only if $\frac{a_i}{b_i}=k$ for some constant $k\in\mathbb{R}^+$, for all $1\le i\le n$ which have $a_ib_i \neq 0$. $_\square$

Given real numbers $a,$ $b,$ and $c$ such that $a^2+b^2+c^2=1,$ what is the maximum value of $12a+4b+3c?$

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Notice that the given equation contains a sum of squares and the expression to maximize is a sum of products. These expressions can be related through the Cauchy-Schwarz inequality:

$$\begin{aligned} \big(12^2+4^2+3^2\big)\big(a^2+b^2+c^2\big) &\ge (12a+4b+3c)^2 \\ (169)(1) &\ge (12a+4b+3c)^2 \\\\ \Rightarrow -13 &\le 12a+4b+3c \le 13. \end{aligned}$$

Thus, the maximum value of $12a+4b+3c$ is $\boxed{13}.$

Furthermore, this maximum is achieved only if

$$\frac{a}{12}=\frac{b}{4}=\frac{c}{3}.$$

Solving this system of equations gives $a=\frac{12}{13},$ $b=\frac{4}{13},$ and $c=\frac{3}{13}.$ $_\square$

Main Article: Triangle Inequality

The triangle inequality relates the lengths of the sides of a triangle.

Let $a,$ $b,$ and $c$ be the side lengths of a triangle, with $a\le b \le c.$ Then,

$$a+b>c.$$

7 and 11 are the lengths of the sides of a triangle. What is the interval of lengths the third side could have?

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Let the length of the third side be $x.$

Case 1: The third side is the longest side.
By the triangle inequality,

$$\begin{aligned} 7+11 &> x \\ 18 &> x. \end{aligned}$$

Case 2: The third side is not the longest side.
Then 11 must be the length of the longest side. By the triangle inequality,

$$\begin{aligned} 7+x &> 11 \\ x &> 4. \end{aligned}$$

The interval of lengths the third side could have is $(4,18).$ $_\square$

Convex and concave functions defined over a closed interval always have a maximum or minimum (respectively) at one of the endpoints of the interval.

Given a convex continuous function $f(x)$ defined over the interval $[a,b],$

$$f(x) \le \max\big(f(a),f(b)\big) \text{ for all }x \in [a,b].$$

The maximum value of this convex function is $f(b)$

Similarly, given a concave continuous function $f(x)$ defined over the interval $[a,b],$

$$f(x) \ge \min\big(f(a),f(b)\big) \text{ for all }x \in [a,b].$$

The minimum value of this concave function is $f(a)$

These relationships are generalized further with Jensen's inequality.

Main Article: Power Mean Inequality (QAGH)

The QM-AM-GM-HM inequality, sometimes known as the QAGH inequality, generalizes a relationship between different types of means (quadratic mean, arithmetic mean, geometric mean, and harmonic mean). It follows from the AM-GM inequality and the Cauchy-Schwarz inequality.

Given a list of $k$ positive real numbers $a_1, \ldots, a_k$, let $f_{\text{QM}}$ denote the quadratic mean, $f_{\text{AM}}$ the arithmetic mean, $f_{\text{GM}}$ the geometric mean, and $f_{\text{HM}}$ the harmonic mean. Then

$$f_{\text{QM}} \geq f_{\text{AM}} \geq f_{\text{GM}} \geq f_{\text{HM}}.$$

Furthermore, equality is achieved if and only if $a_1 = \cdots = a_k.$

If $x,$ $y,$ and $z$ are positive real numbers such that $x^2+4y^2+9z^2=27,$ then what is the minimum value of $\frac{1}{x}+\frac{1}{2y}+\frac{1}{3z} ?$

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Notice that the equation contains a sum of squares, and the expression to be minimized contains a sum of reciprocals. These expressions can be written in terms of the quadratic mean and the harmonic mean, respectively. Thus, they can be related by the QAGH inequality:

$$\begin{aligned} \sqrt{\frac{x^2+4y^2+9z^2}{3}} & \ge \frac{3}{\frac{1}{x}+\frac{1}{2y}+\frac{1}{3z}} \\ 3 & \ge \frac{3}{\frac{1}{x}+\frac{1}{2y}+\frac{1}{3z}}\qquad & (\text{Note: The expression in the denominator is positive.})\\ \frac{1}{x}+\frac{1}{2y}+\frac{1}{3z} & \ge 1. \end{aligned}$$

The minimum value of $\frac{1}{x}+\frac{1}{2y}+\frac{1}{3z}$ is $\boxed{1}.$

Furthermore, this minimum is achieved only if

$$x=2y=3z.$$

Solving this system of equations gives $x=3,$ $y=\frac{3}{2},$ and $z=1.$ $_\square$

The power mean inequality generalizes a relationship between the roots of sums of powers.

For positive $a_1, \ldots, a_k,$ with $f_n$ defined as

$$f_n = \sqrt[n]{\frac{a_1^n + \cdots +a_k^n}{k}},$$

if $m > n,$ it follows that

$$f_m \geq f_n,$$

with equality holding if and only if $a_1 = \cdots = a_k.$

Given that $x$ and $y$ are positive real numbers such that $x+\frac{3}{x}+\frac{1}{x^2}+y^2+3y+\frac{1}{y}=48,$ what is the maximum value of $x^{2/3}+2x^{-1/3}+x^{-4/3}+y^{4/3}+2y^{1/3}+y^{-2/3} ?$

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The expression to be maximized may look intimidating, but careful observation leads to expressing it as a sum of squares:

$$x^{2/3}+2x^{-1/3}+x^{-4/3}+y^{4/3}+2y^{1/3}+y^{-2/3} = \left(x^{1/3}+x^{-2/3}\right)^2 + \left(y^{2/3}+y^{-1/3}\right)^2.$$

Cubing these binomials leads to an expression that is nearly identical to the expression in the given equation:

$$\begin{aligned} \left(x^{1/3}+x^{-2/3}\right)^3 + \left(y^{2/3}+y^{-1/3}\right)^3 &= x+3+\frac{3}{x}+\frac{1}{x^2}+y^2+3y+3+\frac{1}{y} \\ &=54. \end{aligned}$$

Now the sum of squares and the sum of cubes can be related by the power mean inequality:

$$\begin{aligned} \sqrt[3]{\frac{\left(x^{1/3}+x^{-2/3}\right)^3 + \left(y^{2/3}+y^{-1/3}\right)^3}{2}} & \ge \sqrt{\frac{\left(x^{1/3}+x^{-2/3}\right)^2 + \left(y^{2/3}+y^{-1/3}\right)^2}{2}} \\ \\ \sqrt[3]{\frac{54}{2}} & \ge \sqrt{\frac{x^{2/3}+2x^{-1/3}+x^{-4/3}+y^{4/3}+2y^{1/3}+y^{-2/3}}{2}} \\ \\ 18 & \ge x^{2/3}+2x^{-1/3}+x^{-4/3}+y^{4/3}+2y^{1/3}+y^{-2/3}. \end{aligned}$$

Thus, the maximum value of the expression is $\boxed{18}.$ This maximum is achieved only if

$$x^{1/3}+x^{-2/3}=y^{2/3}+y^{-1/3}.$$

The values of $x$ and $y$ that maximize the expression are not easily solved for by hand. The use of computer software yields multiple possible ordered pairs $(x,y)$ that maximize the expression. $_\square$

Main Article: Schur's Inequality

Schur's inequality relates three non-negative real numbers.

For non-negative real numbers $x,$ $y,$ and $z,$ and for positive real number $t$,

$$x^t(x-y)(x-z)+y^t(y-x)(y-z)+z^t(z-x)(z-y) \ge 0,$$

with equality if and only if $x=y=z$, or if two of $x$, $y$, $z$ are equal and the third is 0.

When $t=1,$ the following special case arises:

$$x^3+y^3+z^3+3xyz \ge xy(x+y)+xz(x+z)+yz(y+z).$$

Main Article: Jensen's Inequality

Jensen's inequality is a generalization about the values of convex and concave functions.

Let a real valued function $f$ be convex on the interval $I$. Let $x_1,...,x_n\in I$ and $\omega_1,...,\omega_n\ge 0$. Then we have

$$\dfrac{\omega_1 f\left(x_1\right)+\omega_2 f\left(x_2\right)+\cdots+\omega_n f\left(x_n\right)}{\omega_1+\omega_2+\cdots+\omega_n} \ge f\left(\dfrac{\omega_1x_1+\omega_2x_2+\cdots+\omega_nx_n}{\omega_1+\omega_2+\cdots+\omega_n}\right).$$

If $f$ is concave, the direction of inequality is flipped.

In particular, if we take weights $\omega_1=\omega_2=\cdots=\omega_n=1$, we get the inequality

$$\dfrac{f\left(x_1\right)+f\left(x_2\right)+\cdots+f\left(x_n\right)}{n} \ge f\left(\dfrac{x_1+x_2+\cdots+x_n}{n}\right).$$

The function $f(x)$ has the following properties:

• it is continuous and convex function on the interval $[2,11];$
• $f(2)=6;$
• $f(5)=3;$
• $f(11)=15.$

Use Jensen's inequality to provide an upper bound for the possible values of $f(6).$

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Using the special case of Jensen's inequality with all weights equal to $1,$

$$\begin{aligned} \frac{f(2)+f(5)+f(11)}{3} &\ge f \left(\frac{2+5+11}{3}\right) \\ \\ 8 &\ge f(6). \end{aligned}$$

Thus, an upper bound for the possible values of $f(6)$ is $8.\ _\square$

Graphing the points suggests that a lesser upper bound could be found:

Main Article: Young's Inequality

Young's inequality is a special case of the weighted AM-GM inequality.

Let $p,q$ be positive real numbers satisfying $\frac1{p} + \frac1{q} = 1.$ Then if $a,b$ are nonnegative real numbers,

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

and equality holds if and only if $a^p=b^q.$

Given $x$ and $y$ are positive real numbers such that $\frac{1}{x}+\frac{1}{y}=1,$ what is the minimum value of $\frac{2^x}{x}+\frac{3^y}{y}?$

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These expressions can be related directly with Young's inequality:

$$(2)(3) \le \frac{2^x}{x}+\frac{3^y}{y}.$$

Thus, the minimum value of the expression is $\boxed{6}.$ Furthermore, this minimum is achieved only if

$$2^x=3^y.$$

Solving this system of equations gives $x=\log_2{6}$ and $y=\log_3{6}.$ $_\square$

Main Article: Hölder's Inequality

Hölder's inequality acts as a generalization of the Cauchy-Schwarz inequality.

For sequences $\{a_i\}, \{b_i\}, \ldots, \{z_i\}$, the inequality

$$(a_1+a_2+\cdots+a_n)^{\lambda_a}\cdots(z_1+z_2+\cdots+z_n)^{\lambda_z} \geq a_1^{\lambda_a}b_1^{\lambda_b}\cdots z_1^{\lambda_z}+\cdots+a_n^{\lambda_a}b_n^{\lambda_b}\cdots z_n^{\lambda_z}$$

holds for all $\lambda_a+\lambda_b+\cdots+\lambda_z=1$. For instance, in the case of $\lambda_a=\lambda_b=\frac{1}{2}$, Hölder's inequality reduces to

$$(a_1+a_2+\cdots+a_n)^{\frac{1}{2}}(b_1+b_2+\cdots+b_n)^{\frac{1}{2}} \geq (a_1b_1)^{\frac{1}{2}}+(a_2b_2)^{\frac{1}{2}}+\cdots+(a_nb_n)^{\frac{1}{2}},$$

which is the Cauchy-Schwarz inequality.

Unlike some other classical inequalities, Hölder's inequality does not guarantee a maximum or minimum value, merely an upper bound or lower bound.

Given real numbers $a$ and $b$ such that $a^6+b^6=64,$ use Hölder's inequality to establish an upper bound for the expression $3a^2b+4ab^2.$

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Relate the expressions through Hölder's inequality:

$$\begin{aligned} \big(3^2+4^2\big)^\frac{1}{2} \big(a^6+b^6\big)^\frac{1}{3} \big(b^6+a^6\big)^\frac{1}{6} &\ge 3a^2b+4ab^2 \\ 25^\frac{1}{2} \cdot 64^\frac{1}{3} \cdot 64^\frac{1}{6} &\ge 3a^2b+4ab^2 \\ 40 &\ge 3a^2b+4ab^2. \end{aligned}$$

Thus, an upper bound for the expression is $\boxed{40}.$ Using computer software, the actual maximum value of the expression is $\approx 39.6217,$ when $a \approx 1.76652$ and $b \approx 1.79645.$ $_\square$

Minkowski's inequality follows from Hölder's inequality:

$$\left(\sum _{ n=1 }^{ k } ({ x }_{ n }+{ y }_{ n })^p\right)^{ \frac { 1 }{ p } }\le \left(\sum _{ n=1 }^{ k }{ { x }_{ n }^{ p } } \right) ^{ \frac { 1 }{ p } }+\left(\sum _{ n=1 }^{ k }{ { y }_{ n }^{ p } } \right)^{ \frac { 1 }{ p } }$$

for $p>1$ and ${x}_{n},{y}_{n}\ge0.$

Main Article: Rearrangement Inequality

The rearrangement inequality relates sums of products of elements of non-increasing sequences of real numbers.

Given two sequences $a_1 \geq a_2 \geq \cdots \geq a_n$ and $b_1 \geq b_2 \geq \cdots \geq b_n$,

$$a_1b_1+a_2b_2+\cdots+a_nb_n \geq a_1b_{\pi(1)}+a_2b_{\pi(2)}+\cdots+a_nb_{\pi(n)} \geq a_1b_n+a_2b_{n-1}+\cdots+a_nb_1,$$

where $\pi(1), \pi(2), \ldots, \pi(n)$ is any permutation of $1, 2, \ldots, n$.

Main Article: Reverse Rearrangement Inequality

The reverse rearrangement inequality relates products of sums of elements of non-increasing sequences of real numbers. One can see its similarity to the rearrangement inequality; the operations of addition and multiplication are merely exchanged, with the inequality signs being inverted as well.

Given two non-negative sequences of real numbers $\{a_n\}$ and $\{b_n\}$ that are similarly ordered,

$$\prod_{k=1}^n(a_k+b_k)\le \prod_{k=1}^n \big(a_k+b_{\sigma(k)}\big)\le \prod_{k=1}^n (a_k+b_{n-k+1}),$$

where $\sigma(1),\sigma(2),\ldots, \sigma(n)$ are permutations of $1,2,\ldots, n,$ respectively. $_\square$

Main Article: Chebyshev's Inequality

Chebyshev's inequality is an extension of the rearrangement inequality. It is a generalization of the relationship between non-increasing sequences of real numbers.

Given non-increasing sequences of real numbers $\{a_i\}$ and $\{b_i\},$

$${\frac {\sum _{ i=1 }^{ n }{ { a }_{ i }{ b }_{ i } } }{ n } \ge \frac {\sum _{ i=1 }^{ n }{ { a }_{ i } } }{ n } \times \frac {\sum _{ i=1 }^{ n }{ { b }_{ i } } }{ n } \ge \frac {\sum _{ i=1 }^{ n }{ { a }_{ i }{ b }_{ n+1-i } } }{ n } }.$$

Main Article: Muirhead's Inequality

Suppose $\{a_i\}$ and $\{b_i\}$ are sequences of real numbers such that $\{a_i\}$ majorizes $\{b_i\}.$ That is,

• $\{a_i\}$ and $\{b_i\}$ are non-increasing, each with $n$ non-negative terms;
• $\sum\limits_{j=1}^{k}a_j \ge \sum\limits_{j=1}^{k}b_j,$ for all integers $k$ such that $1 \le k < n;$
• $\sum\limits_{j=1}^{n}a_j = \sum\limits_{j=1}^{n}b_j.$

Then for all non-negative $x_i,$

$$\sum_{\text{sym}} x_1^{a_1}\cdots x_n^{a_n} \ge \sum_{\text{sym}} x_1^{b_1}\cdots x_n^{b_n}.$$

Suppose $x,$ $y,$ and $z$ are non-negative real numbers. Use Muirhead's inequality to show that $x^5+y^5+z^5 \ge x^3yz+xy^3z+xyz^3.$

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Note that both sides of the inequality are symmetric polynomials of the same degree. The non-increasing sequence associated with the left-hand side of the inequality is $\{5,0,0\}.$ The non-increasing sequence associated with the right-hand side of the inequality is $\{3,1,1\}.$ It can be shown that the left sequence majorizes the right sequence:

$$\begin{aligned} 5 &> 3 \\ 5+0 &> 3+1 \\ 5+0+0 &= 3+1+1. \end{aligned}$$

Since the conditions of Muirhead's inequality are satisfied, the following inequality is true for all non-negative $x,$ $y,$ and $z:$

$$x^5+y^5+z^5 \ge x^3yz+xy^3z+xyz^3.\ _\square$$

• Newton's Inequality

• Maclaurin's Inequality

• Popoviciu's Inequality

• Bernoulli's Inequality