Lesson Arithmetic mean and geometric mean inequality
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<H2>Arithmetic mean and geometric mean inequality</H2> The Arithmetic mean - Geometric mean inequality is a famous, classic and basic Theorem on inequalities. It states that . . . <BLOCKQUOTE><H3>AM-GM Theorem</H3>Geometric mean of two real positive numbers is lesser than or equal to their arithmetic mean. Geometric mean of two real positive unequal numbers is less than their arithmetic mean. Geometric mean of two real positive numbers is equal to their arithmetic mean if and only if the two numbers are equal.</BLOCKQUOTE> <B>Proof</B> Let <B>a</B> and <B>b</B> be two real positive numbers. Their arithmetic mean is {{{(a + b)/2}}}. Their geometric mean is {{{sqrt(a*b)}}}. The <B>Theorem</B> states and we need to prove that {{{sqrt(a*b)}}} <= {{{(a + b)/2}}}. Let us <B><U>start</U></B> with inequality {{{(sqrt(a) - sqrt(b))^2}}} >= 0. (1) This inequality is always true because the square of a real number is non-negative. Now, expand (1) as {{{(sqrt(a) - sqrt(b))^2}}} = {{{(sqrt(a))^2}}} - {{{2*sqrt(a)*sqrt(b)}}} + {{{(sqrt(b))^2}}} = {{{a}}} - {{{2*sqrt(a)*sqrt(b)}}} + {{{b}}}. using the formula on a square of the difference. Thus the original inequality (1) is equivalent to {{{a}}} - {{{2*sqrt(a)*sqrt(b)}}} + {{{b}}} >= {{{0}}}, or {{{a}}} + {{{b}}} >= {{{2*sqrt(a)*sqrt(b)}}}, or {{{(a + b)/2}}} >= {{{sqrt(a*b)}}}. (2) It is exactly the <B><U>first</U></B> statement of the Theorem. So, it is proved now. <B><U>Next</U></B>, if two real numbers <B>a</B> and <B>b</B> are unequal, then the inequality (1) is a strong inequality {{{(sqrt(a) - sqrt(b))^2}}} > 0, (3) because in this case the number {{{sqrt(a) - sqrt(b)}}} is non-zero and a square of a non-zero real number is positive. Then the same chain of arguments leads us to the strong inequality similar to (2): {{{(a + b)/2}}} > {{{sqrt(a*b)}}}. (4) It is the <B><U>second</U></B> statement of the Theorem. and it is proved now. <B><U>Finally</U></B>, the <B><U>third</U></B> statement of the Theorem is obvious in one and the other directions in the light of the used arguments. <H3>Example 1</H3>Let <B>a</B> = 2, <B>b</B> = 8. Calculate and compare Arithmetic mean and Geometric mean of these numbers. <B>Solution</B> Arithmetic mean of numbers 2 and 8 is {{{(2+8)/2}}} = {{{5}}}. Geometric mean of these numbers is {{{sqrt(2*8)}}} = {{{sqrt(16)}}} = {{{4}}}. The Geometric mean is lesser than the Arithmetic mean: {{{4}}} < {{{5}}}. <H3>Example 2</H3>Let <B>a</B> = 4, <B>b</B> = 5. Calculate and compare Arithmetic mean and Geometric mean of these numbers. <B>Solution</B> Arithmetic mean of numbers 4 and 5 is {{{(4+5)/2}}} = {{{4.5}}}. Geometric mean of these numbers is {{{sqrt(4*5)}}} = {{{sqrt(20)}}} = {{{4.472}}} (approximately). The Geometric mean is lesser than the Arithmetic mean: {{{4.472}}} < {{{4.5}}}. <H3>Example 3</H3>Let <B>a</B> = 5, <B>b</B> = 5. Calculate and compare Arithmetic mean and Geometric mean of these numbers. <B>Solution</B> Arithmetic mean of numbers 5 and 5 is {{{(5+5)/2}}} = {{{5}}}. Geometric mean of these numbers is {{{sqrt(5*5)}}} = {{{sqrt(25)}}} = {{{5}}} (approximately). It is the case when the Geometric mean is equal to the Arithmetic mean: {{{5}}} = {{{5}}}. <H3>Example 4</H3>What number is greater, {{{sqrt(7.3*6.7)}}} or {{{7}}} ? <B>Solution</B> Notice that the Arithmetic mean of numbers 6.7 and 7.3 is {{{(6.7+7.3)/2}}} = {{{7}}}. In accordance with AM-GM Theorem, {{{sqrt(6.7*7.3)}}} < {{{7}}}. Notice that we got the conclusion without calculating the product {{{6.7*7.3}}}. <H3>Example 5</H3>What number is greater, {{{51*53}}} or {{{52^2}}} ? <B>Solution</B> Notice that the Arithmetic mean of numbers 51 and 53 is {{{(51+53)/2}}} = {{{52}}}. In accordance with <B>AM-GM Theorem</B>, {{{sqrt(51*53)}}} < {{{52}}}. Hence, {{{51*53}}} < {{{52^2}}}. Again, we were able to get the conclusion without calculating the product {{{51*53}}} and {{{52^2}}}. <H3>Example 6</H3>What number is greater, {{{99*101}}} or {{{100^2}}} ? <B>Solution</B> Notice that the Arithmetic mean of numbers 99 and 101 is {{{(51+53)/2}}} = {{{52}}}. The same arguments as in the <B>Example 5</B> above lead us to the conclusion that {{{99*101}}} < {{{100^2}}}. Surely, you can notice that 99 = 100-1, 101 = 100+1 and to write {{{99*101}}} = {{{(100-1)*(100+1)}}} = {{{100^2}}} - {{{1^2}}} = {{{100^2}}} - {{{1}}} < {{{100^2}}}. It gives you another proof of the <B>Example 6</B> inequality. The next lesson provides a geometric interpretations to the Arithmetic mean and geometric mean inequality - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Arithmetic-mean-and-geometric-mean-inequality-Geometric-interpretations.lesson>Arithmetic mean and geometric mean inequality - Geometric interpretations</A> My other lessons on solving inequalities are - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Solving-simple-and-simplest-inequalities.lesson>Solving simple and simplest linear inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Solving-absolute-value-inequalities-IK.lesson>Solving absolute value inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Advanced-problems-on-solving-absolute-value-inequalities.lesson>Advanced problems on solving absolute value inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Solving-systems-of-linear-inequalities-in-one-unknown.lesson>Solving systems of linear inequalities in one unknown</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Solving-compound-inequalities.lesson>Solving compound inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Which-number-is-greater-Comparing-magnitude-of-irrational-numbers.lesson>What number is greater? Comparing magnitude of irrational numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Arithmetic-mean-and-geometric-mean-inequality-Geometric-interpretations.lesson>Arithmetic mean and geometric mean inequality - Geometric interpretations</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Harmonic-mean.lesson>Harmonic mean</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Prove-that-if-a-b-and-c-are-the-sides-of-a-triangle-then-so-are-sqrt%28a%29-sqrt%28b%29-and-sqrt%28c%29.lesson>Prove that if a, b, and c are the sides of a triangle, then so are sqrt(a), sqrt(b), and sqrt(c)</A> - <A HREF=http://www.algebra.com/algebra/homework/Inequalities/Solving-problems-on-quadratic-inequalities.lesson>Solving problems on quadratic inequalities</A> - <A HREF=http://www.algebra.com/algebra/homework/Inequalities/Solving-inequalities-for-polynomials-factored-to-a-product-of-linear-binomials.lesson>Solving inequalities for high degree polynomials factored into a product of linear binomials</A> - <A HREF=http://www.algebra.com/algebra/homework/Inequalities/Solving-inequalities-for-rat-functions-with-num-and-denom-factored-into-a-product-of-linear-binomials.lesson>Solving inequalities for rational functions with numerator and denominator factored into a product of linear binomials</A> - <A HREF=http://www.algebra.com/algebra/homework/Inequalities/Solving-inequalities-for-rational-functions-with-non-zero-right-side.lesson>Solving inequalities for rational functions with non-zero right side</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Another-way-solving-inequalities-for-rational-functions-with-non-zero-right-siode.lesson>Another way solving inequalities for rational functions with non-zero right side</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Advanced-lesson-on-inequalities.lesson>Advanced problems on inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Challenging-problems-on-inequalities.lesson>Challenging problems on inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Graphs/Solving-systems-of-inequalities-in-two-unknown-graphically-in-a-coordinate-plane.lesson>Solving systems of inequalities in two unknowns graphically in a coordinate plane</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Solving-word-problems-on-inequalities.lesson>Solving word problems on inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Proving-inequalities.lesson>Proving inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Math-circle-level-problem-on-inequalities.lesson>Math circle level problem on inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Math-Olympiad-level-problem-on-inequalities.lesson>Math Olympiad level problems on inequalities</A> - <A HREF=https://www.algebra.com/algebra/homework/Inequalities/Enterteinment-problems-on-inequalities.lesson>Entertainment problems on inequalities</A> under the topic <B>Inequalities, trichotomy</B> of the section <B>Algebra-I</B>. My lessons on domains of functions are - <A HREF=http://www.algebra.com/algebra/homework/Functions/Finding-domain-of-a-function.lesson>Domain of a function which is a quadratic polynomial under the square root operator</A> - <A HREF=http://www.algebra.com/algebra/homework/Functions/Domain-of-a-function-containing-high-degree-polynomial-under-the-square-root.lesson>Domain of a function which is a high degree polynomial under the square root operator</A> - <A HREF=http://www.algebra.com/algebra/homework/Functions/Domain-of-a-function-which-is-the-square-root-of-a-rat-function.lesson>Domain of a function which is the square root of a rational function</A>. under the topic <B>Functions, Domain</B> of the section <B>Algebra-I</B>. See also <A HREF=http://www.algebra.com/algebra/homework/Inequalities/OVERVIEW-of-lessons-on-inequalities-and-domains-of-functions.lesson>OVERVIEW of lessons on inequalities and domains of functions</A>. Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
