AM GM Inequality Solved Problems

In Mathematics, inequality is a relation that makes a non-equal comparison between mathematical expressions or two numbers. The AM–GM inequality, or inequality of arithmetic and geometric means, states that the arithmetic means of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list.

If every number in the list is the same, only then there is a possibility that two means are equal. In this section, we will practise problems related to AM–GM inequality.

The simplest non-trivial case, i.e., with more than one variable for two non-negative numbers x and y, is the statement that

\begin{array}{l} \frac{x + y}{2} \geq \sqrt{x y} \end{array}

with equality if and only if x = y. This case can be seen from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the elementary case

\begin{array}{l} (a \pm b)^{2} = a^{2} \pm 2 a b + b^{2} \end{array}

of the binomial formula:

\begin{array}{l} \begin{aligned} 0 & \leq (x - y)^{2} \\ = x^{2} - 2 x y + y^{2} \\ = x^{2} + 2 x y + y^{2} - 4 x y \\ = (x + y)^{2} - 4 x y . \end{aligned} \end{array}

Hence, (x + y)² ≥ 4xy, with equality precisely when (x – y)² = 0, i.e., x = y.

The AM–GM inequality then follows by taking the positive square root of both sides.

For a geometrical interpretation, consider a rectangle with sides of length x and y, hence it has a perimeter 2x + 2y and area xy. Similarly, a square with all sides of length √(xy) has a perimeter of 4√(xy) and the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that

\begin{array}{l} 2 x + 2 y \geq 4 \sqrt{x y} \end{array}

and that only the square has the smallest perimeter amongst all rectangles of equal area.

AM-GM Inequality Relationship

Let a₁, a₂,…, a_n be n positive real numbers. The Arithmetic Mean and Geometric Mean are defined as:

\begin{array}{l} A M = \frac{a_{1} + a_{2} + \dots + a_{n}}{n} \end{array}

\begin{array}{l} G M = \sqrt[n]{a_{1} a_{2} \dots a_{n}} \end{array}

The AM–GM inequality states that

AM ≥ GM

Solved Examples on AM-GM Inequality

Example 1: Prove that

\begin{array}{l} (α + β + γ) (α β + β γ + γ α) \geq g α β γ \end{array}

Solution: Applying AM > GM

\begin{array}{l} \frac{α + β + γ}{3} \geq {(α β γ)}^{^{1} /_{3}} \dots \dots \dots (1) \end{array}

\begin{array}{l} \frac{α β + β γ + γ α}{3} \geq {(α^{2} β^{2} γ^{2})}^{^{1} /_{3}} \dots \dots \dots (2) \end{array}

On multiplying equations (1) and (2),

We get

\begin{array}{l} \frac{α + β + γ (α β + β γ + γ α)}{g} \geq {(α β γ)}^{^{1} /_{3}} {(α β γ)}^{^{1} /_{3}} \geq (α β γ) \end{array}

Therefore,

\begin{array}{l} (α + β + γ) (α β + β γ + γ α) \geq g α β γ \end{array}

Hence proved.

Example 2: Insert 3 GMs between 1 and 16

Solution: Let the terms be 1, g₁, g₂, g₃, 16

\begin{array}{l} r = {(\frac{16}{1})}^{\frac{1}{3 + 1}} \end{array}

\begin{array}{l} r = {(16)}^{1 / 4} = 2 \end{array}

Three geometric means between 1 and 16 are 2, 4, 8.

Example 3: If P, Q, R are positive real numbers such that P + Q + R = g, find the maximum value of P²Q³R⁴.

Solution: P + Q + R = g

Applying AM > GM

\begin{array}{l} 2 (\frac{P}{2}) + 3 (\frac{Q}{3}) + 4 (\frac{R}{4}) = g \end{array}

\begin{array}{l} \Rightarrow ({(\frac{P}{2})}^{2} . {(\frac{Q}{3})}^{3} . {(\frac{R}{4})}^{4}) \leq \frac{g}{3} \end{array}

3g, 22, 33, 44 > P²Q²R⁴

\begin{array}{l} 3^{12} \cdot 2^{10} \geq P^{2} Q^{3} R^{4} \end{array}

\begin{array}{l} 6^{10} \cdot 3^{2} \geq P^{2} Q^{3} R^{4} \end{array}

Example 4: Evaluate 1² + 2² + 3² + 4² + 5².

Solution: Above expression can be written as

\begin{array}{l} \sum_{m = 1}^{5} m^{2} \end{array}

We know, sum of squares of first five natural numbers = 1/6 {n(n+1)(2n+1)}

Here, n = 5.

Therefore,

\begin{array}{l} \sum_{m = 1}^{5} m^{2} \end{array}

= 5(5+1)(2.5+1)/6 = 55

Example 5: Prove (a + b + c) (ab + bc + ca) > 9abc.

Solution:

\begin{array}{l} \frac{a + b + c}{3} > (a b c)^{1 / 3} \end{array}

and

\begin{array}{l} \frac{a b + b c + c a}{3} > (a b . b c . c a)^{\frac{1}{3}} \end{array}

[Using AM > GM]

Multiplying the above results, we get

\begin{array}{l} \frac{a + b + c}{3} \times \frac{a b + b c + c a}{3} > (a b c)^{1 / 3} (a b . b c . c a)^{1 / 3} \end{array}

= (a³b³c³)^1/3

= abc

⇒ (a + b + c)(ab + bc + ca) > 9abc.

Hence proved.

Example 5: Consider the function

\begin{array}{l} f (x, y, z) = \frac{x}{y} + \sqrt{\frac{y}{z}} + \sqrt[3]{\frac{z}{x}} \end{array}

for all positive real numbers

x

,

y

and

z

. Find the minimal value of the function.

Solution:

\begin{array}{l} \begin{aligned} f (x, y, z) & = 6 \cdot \frac{\frac{x}{y} + \frac{1}{2} \sqrt{\frac{y}{z}} + \frac{1}{2} \sqrt{\frac{y}{z}} + \frac{1}{3} \sqrt[3]{\frac{z}{x}} + \frac{1}{3} \sqrt[3]{\frac{z}{x}} + \frac{1}{3} \sqrt[3]{\frac{z}{x}}}{6} \\ = 6 \cdot \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}}{6} \end{aligned} \\ w i t h x_{1} = \frac{x}{y}, x_{2} = x_{3} = \frac{1}{2} \sqrt{\frac{y}{z}}, x_{4} = x_{5} = x_{6} = \frac{1}{3} \sqrt[3]{\frac{z}{x}} . \end{array}

Applying the AM–GM inequality for $n = 6$ , we get

\begin{array}{l} \begin{aligned} f (x, y, z) & \geq 6 \cdot \sqrt[6]{\frac{x}{y} \cdot \frac{1}{2} \sqrt{\frac{y}{z}} \cdot \frac{1}{2} \sqrt{\frac{y}{z}} \cdot \frac{1}{3} \sqrt[3]{\frac{z}{x}} \cdot \frac{1}{3} \sqrt[3]{\frac{z}{x}} \cdot \frac{1}{3} \sqrt[3]{\frac{z}{x}}} \\ = 6 \cdot \sqrt[6]{\frac{1}{2 \cdot 2 \cdot 3 \cdot 3 \cdot 3} \frac{x}{y} \frac{y}{z} \frac{z}{x}} \\ = 2^{2 / 3} \cdot 3^{1 / 2} . \end{aligned} \end{array}

Further, we know that the two sides are equal exactly when all the terms of the mean are equal.

\begin{array}{l} f (x, y, z) = 2^{2 / 3} \cdot 3^{1 / 2} when \frac{x}{y} = \frac{1}{2} \sqrt{\frac{y}{z}} = \frac{1}{3} \sqrt[3]{\frac{z}{x}} . \end{array}

All the points (x, y, z), satisfying these conditions lie on a half-line starting at the origin and are given by,

\begin{array}{l} (x, y, z) = (t, \sqrt[3]{2} \sqrt{3} t, \frac{3 \sqrt{3}}{2} t) w i t h t > 0. \end{array}

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