Classroom. The Arithmetic Mean Geometric Mean Harmonic Mean: Inequalities and a Spectrum of Applications

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1 Classroom In this section of Resonance, we invite readers to pose questions likely to be raised in a classroom situation. We may suggest strategies for dealing with them, or invite responses, or both. Classroom is equally a forum for raising broader issues and sharing personal experiences and viewpoints on matters related to teaching and learning science. The Arithmetic Mean Geometric Mean Harmonic Mean: Inequalities and a Spectrum of Applications The Arithmetic Mean Geometric Mean Harmonic Mean inequality, AM GM HM inequality in short, is one of the fundamental inequalities in Algebra, and it is used extensively in olympiad mathematics to solve many problems. The aim of this article is to acquaint students with the inequality, its proof and various applications. Before we state the AM GM HM inequality, let us define the Arithmetic Mean (AM), Geometric Mean (GM) and Harmonic Mean (HM). Let a, a,...,a n be n positive real numbers. The AM (A n ), GM (G n ) and HM (H n )ofthesen numbers are defined as: Prithwijit De Homi Bhabha Centre for Science Education Tata Institute of Fundamental Research Mumbai prithwijit@hbcse.tifr.res.in AM GM HM inequality is one of the fundamental inequalities in Algebra. A n = a + a + + a n, () n G n = n a a a n, () n H n = (3) a a a n The AM GM HM inequality states that A n G n H n, (4) Keywords Arithmetic mean, geometric mean, harmonic mean, Nesbit s inequality, Euler s inequality. RESONANCE December 06 9

2 The AM GM HM inequality states that A n G n H n,and equality occurs if and only if a = a = = a n. and equality occurs if and only if a = a = = a n. The result is easy to prove when n =. In this case we need to show, a + a a a +. (5) a a Observe that a + a a a = ( a a ) 0. (6) This proves that A G. To prove G H observe that writing b = /a and b = /a reduces the relevant inequality to b + b b b (7) To prove G n H n for any n is same as proving A n G n for the reciprocals of the given real numbers. which we have already proved. This proves the AM GM HM inequality for n =. Note that to prove G n H n for any n is same as proving A n G n for the reciprocals of the given real numbers. Thus to prove the AM GM HM inequality it suffices to prove that A n G n for all n. Forn = 3 the inequality can be proved by using elementary algebra. We have to prove a + b + c 3 This is same as proving, 3 abc. (8) (a + b + c) 3 7abc. (9) But, (a + b + c) 3 = a 3 + b 3 + c 3 + 3(a + b)(b + c)(c + a). (0) Therefore, (a + b + c) 3 7abc = (a 3 + b 3 + c 3 3abc) + 3{(a + b)(b + c)(c + a) 8abc}. () 0 RESONANCE December 06

3 We can write, a 3 + b 3 + c 3 3abc = (a + b) 3 + c 3 3ab (a + b + c) = (a + b + c) {(a + b) and further simplify it to, (a + b) c + c } 3ab (a + b + c), () a 3 +b 3 +c 3 3abc = (a+b+c)(a +b +c ab bc ca). (3) But, a +b +c ab bc ca = {(a b) +(b c) +(c a) }. (4) Thus, a 3 +b 3 +c 3 3abc = (a+b+c){(a b) +(b c) +(c a) } 0. (5) Using the AM GM HM inequality for n = we see that, (a + b )(b + c)(c + a) ( ab)( bc)( ca) = 8abc. (6) Therefore, (a + b + c) 3 7abc = (a 3 + b 3 + c 3 3abc) + 3{(a + b)(b + c)(c + a) 8abc } 0, (7) and we have proved A 3 G 3. Observe that A 3 = G 3 if and only if equality occurs in (5) and (6) which is same as saying a = b = c. One may continue in the same spirit to prove the inequality for higher values of n but would realize during the process that the algebra becomes cumbersome and would almost certainly give up the chase. The case n = 4 can be dealt with quite elegantly by a clever reduction to the case n =. How? Observe that, a + a + a 3 + a 4 4 = a + a + a 3 + a 4. (8) RESONANCE December 06

4 Now set b = a + a b a 3 a 4 and and b = a 3 + a 4. Then b a a, a + a + a 3 + a 4 = b + b b b = 4 a a a 3 a 4. (9) Hence, a + a + a 3 + a 4 4 a a a 3 a 4, (0) 4 with equality if and only if a = a = a 3 = a 4. This argument can be easily extended to n = 8 and more generally to any n of the form k for some positive integer k. For instance, when n = 8, apply A 4 G 4 to the set of numbers { a + a, a 3 + a 4, a 5 + a 6, a 7 + a } 8 followed by A G to the sets of numbers {a, a }, {a 3, a 4 }, {a 5, a 6 }, {a 7, a 8 }. Thus for n = k, a + a + + a k k k a a a k. () If n is not a power of two then we can find a unique k such that k < n < k+ and we observe that ( k+ n) times A n = a { }} { + a + + a n + (A n + + A n ) k+ k+ a a a n.(a n ) k+ n. () This can be simplified further to obtain, A n n a a a n, (3) and this is same as A n G n. The idea of this proof is due to the great French mathematician Augustin-Louis Cauchy ( ). There is another way to prove A n G n for all positive integers n. Let, a i x i = n, (4) a a a n RESONANCE December 06

5 for i =,,...,n. Then x x x n = and to prove A n G n we need to show that, x + x + + x n n, (5) AM GM HM inequality is used extensively in olympiad mathematics to solve many problems. subject to x x x n =. When n = it boils down to showing x + x ifx x =. This follows easily because x + x = ( x ) + ( x ) x x = ( x x ) 0. (6) Assume that we have shown that for m 3, x + x + + x m m, (7) if x x x m =. Now x x x m =. Without loss of generality let x x. Thus, (x ) ( x ) 0, (8) implying x + x + x x. (9) Observe that x x x m = = y y y m where y = x x and y i = x i+ for i m. Therefore, y + y + + y m m. (30) But y +y 3 + +y m = x 3 + x x m and +y = + x x x + x. Thus, x + x + + x m m, (3) and by induction it follows that x + x + + x n n for every positive integer n ifx x x n =. Applications Having presented two proofs of the AM GM HM inequality, let us look at some applications.. Let a, a,...,a n be positive real numbers. Let b, b,...,b n be a permutation of a, a,...,a n. Then, RESONANCE December 06 3

6 a b + a b + + a n b n n. In order to prove this just note that a a a n = b b b n. Thus, a b + a b + + a n b n n n a a...a n b b...b n =. (3). (Nesbit s Inequality) Let a, b, c be positive real numbers. Then, a b + c + b c + a + c a + b 3, and equality occurs if and only if a = b = c. Let P = a b + c + b c + a + c a + b. Then, ( P + 3 = (a + b + c) Now A 3 H 3 implies, b + c + c + a + a + b ). (33) a + b + b + c 3 + c + a 3 a + b + b + c + c + a. (34) Therefore P and hence P 3. Equality occurs if and only if A 3 = H 3 which occurs if and only if a + b c + a, i.e., a = b = c. = b + c 3. Let a, b, c be real numbers greater than or equal to. Then, a 3 + b b + + b3 + c c + + c3 + a a + 9. To see this, observe that a 3 + 3(a a+) = a 3 3a +3a = (a ) 3 a 3 + 0, whence a 3. Since every term in the a + = 4 RESONANCE December 06

7 left hand side of the inequality is positive we have, a 3 + b b + + b3 + c c + + c3 + a a + ( (a 3 + ) (b 3 + ) (c 3 ) /3 + ) 3 (a a + ) (b b + ) (c = 9. (35) c + ) 4. Let a, b, c be positive real numbers such that a + b + c =. Then, we have the inequality, (a ) (b ) (c ) + 8 (a + ) (b + ) (c + ) 4. For ( this, observe that a >, b >, c > and a = a b + ) a (by AM GM inequality). Similarly, we c bc get b b and c ca taking the reciprocal we obtain, c ab. Multiplying these and (a ) (b ) (c ) 8. (36) Next observe that a + a = + a whence, a+ a (a ). Similarly we obtain b + (b ) and c + (c ). Multiplying these yields (a + ) (b + ) (c + ) 6 (a ) (b ) (c ) 6.8 = 64. Therefore, 8 (a + ) (b + ) (c + ) 8. (37) By adding (36) and (37) we get, (a ) (b ) (c ) + 8 (a + ) (b + ) (c + ) 4. RESONANCE December 06 5

8 5. (Euler s Inequality) Let R be the circumradius and r be the inradius of a triangle. Then R r with equality if and only if the triangle is equilateral. This is perhaps the most elegant geometric inequality that one comes across in school level mathematics. There is a geometric proof of this result. Here, we shall present an algebraic proof. Let ABC be the given triangle. Denote by a, b, and c the lengths of the sides BC, CA, and AB respectively. Let s = a + b + c, where s is the semi-perimeter, and let Δ denote the area of ABC. Observe that s a, s b and s c are positive quantities. Recall that r = Δ abc and R = s 4Δ. Therefore, R r = abcs 4Δ = abc 4(s a)(s b)(s c). (38) Now observe that, a = s b c = (s b) + (s c) (s b)(s c), (39) b = s c a = (s c) + (s a) (s c)(s a), (40) and, c = s a b = (s a) + (s b) (s a) (s b). (4) Multiplying these three inequalities leads to, whence, abc 8(s a)(s b)(s c) (4) R r = abc. (43) 4(s a)(s b)(s c) Equality occurs if and only if s a = s b = s c, that is, a = b = c, which is same as saying that the triangle is equilateral. 6 RESONANCE December 06

9 Solving Some Equations Using the Inequalities In all the examples discussed so far, we applied the AM GM HM inequality to show that one algebraic expression is at least as large as another algebraic expression under certain conditions on the variables involved. It can also be used to solve algebraic equations. Here are some examples. AM GM HM inequality can be used to solve algebraic equations.. Suppose we wish to find all quadruples a, b, c, d of positive real numbers that are solutions to the system of equations ( a + b + c + d = 4, a + b + c + ) d ( + 3abcd ) = 6. By A 4 G 4 we get, = a + b + c + d 4 4 abcd, (44) a + b + c + d 4 (abcd) 4, (45) Multiplying both sides of (45) by ( + 3abcd ) and simplifying we obtain, ( a + b + c + ) 4(+ 3abcd ) d ( + 3abcd) (abcd ) 3. (46) 4(+ 3t ) Let t = abcd. Then 6 t 3. This implies 4t 3 3t 0. But 4t 3 3t = (t ) (t + ). Therefore (t ) (t + ) 0. Hence t. But earlier we obtained t. Therefore t =. So now we have, a + b + c + d 4 = 4 abcd, that is, A 4 = G 4. Hence a = b = c = d = a + b + c + d =. 4 It is easy to see that (a, b, c, d) = (,,, ) indeed satisfies the given system of equations. RESONANCE December 06 7

10 . Let us solve the simultaneous system of equations for real values of x, y and z. x = y + y ; y = z + z ; z = x + x. Observe that x = y = z = 0 is a solution. If x, y, z are different from zero then they must be positive (Why?). Now observe that for any positive real number t, ( t + ) t. =. (47) t t Therefore, and, y x = ( y + ) y, (48) y z y = ( z + ) z, (49) z x z = ( x + ) x. (50) x Thus x y z x. Hence x = y = z and the common value is. Thus the solutions are (x, y, z) = (0, 0, 0), (,, ). It is easy to see that these triples indeed satisfy the given system. Optimization Problems Some optimization questions can also be solved using the AM GM HM inequalities.. Among all triangles with a fixed perimeter, let us find which triangle has the largest area. Let us also determine the maximum area. 8 RESONANCE December 06

11 Let p be the fixed perimeter. Let a, b and c be the sides of a triangle whose perimeter is p. Let s = p/. The area of the triangle is Δ (a, b, c) = s (s a)(s b)(s c). (5) Some optimization questions can also be solved using the AM GM HM inequalities. Note that the area is a function of the side-lengths a, b and c. Now, (s a) + (s b) + (s c) 3 (s a)(s b)(s c), (5) 3 with equality if and only if s a = s b = s c, that is, a = b = c. Therefore, Δ (a, b, c) = s (s a)(s b)(s c) s 3 3 = p 3. (53) Thus the maximum value of the area is and it is attained 3 when the triangle is equilateral.. Let a sector with central angle θ be removed from a circle of unit radius and rolled into a right circular cone. What is the value of θ for which the volume of the cone thus formed is maximum? Let r and h be respectively the radius of the base and the height of the cone. Then, πr = θ, (54) and, Solving for h to obtain, h = p πr r + h = θ. (55) r = Therefore the volume of the cone is, V = πr h 3 ( θ ). (56) π = π ( θ ) ( θ ) 3. (57) π π RESONANCE December 06 9

12 AM GM HM inequalities can be applied to a system of quadratic equations. Let t = whence, ( θ ). Observe that 0 < t < and, π t + t + ( t) t 3 3 t ( t) (58) t t 3 3. (59) Equality is attained if and only if t = t, that is, t = 3 and this leads to θ = π. Therefore the volume of the cone is 3 maximised when the central angle is equal to π 3 = 94 approximately. Here is an application of the AM GM HM inequalities to a system of quadratic equations. Suppose that the three equations ax bx+c = 0, bx cx+a = 0 and cx ax + b = 0 all have only positive roots. Show that a = b = c. Let α,β be the roots of ax bx + c = 0, α,β be the roots of bx cx + a = 0 and α 3,β 3 be the roots of cx ax + b = 0. Then, α + β = b a ; α β = c a, (60) α + β = c b ; α β = a b, (6) α 3 + β 3 = a c ; α 3β 3 = b c. (6) The roots are positive. By A G we obtain, 30 RESONANCE December 06

13 b a = α + β c b = α + β a c = α 3 + β 3 α β = α β = α 3 β 3 = c a, (63) a b, (64) b c, (65) which are equivalent to b ca, c ab and a bc. Suppose on the contrary, either a, b, c are distinct or any two of them are equal but are different from the third. If a, b, c are distinct then as the inequalities are symmetric with respect to a, b, c, wemay assume without loss of generality that a > b > c. In that case c < ab and we have a contradiction. If a = b > c then also c < ab and we have a contradiction. By symmetry a > b = c and c = a > b will lead to contradictions too. Therefore a = b = c. Weighted AM GM HM Inequalities Let us point out to the reader that for the AM GM inequality to hold, the quantities a, a,...,a n may assumed to be non-negative but they have to be positive for the AM HM inequality or the GM HM inequality to be valid. The AM GM HM inequality can be generailzed to the weighted AM GM HM inequality by assigning weights to the quantities. We state the inequality without proof. Let a, a,...,a n and w, w,...,w n be positive real numbers. Then, n w i a i i= n w i i= (a w aw aw n n w i n ) i= n w i i= n i=. w i a i Here is an application. Let a, b, c be positive real numbers. Prove that RESONANCE December 06 3

14 0a + b + c + a + 0b + c + a + b + 0c 3a + 3b + 3c. By the weighted AM HM inequality we have, 3 0a + b + c 3 a + 0b + c 3 a + b + 0c 0/a + /b + /c, 3 /a + 0/b + /c, 3 /a + /b + 0/c. 3 Upon adding the above inequalities and rearranging terms we obtain, 0a + b + c + a + 0b + c + a + b + 0c 3a + 3b + 3c. We conclude this article by leaving some problems for the student. Problems. Let a, a,...,a n be n positive real numbers whose product is. Prove that ( + a )(+ a ) ( + a n ) n.. Let a, b, c be real numbers greater than or equal to. Prove that a (b + 3) 3c + + b (c + 3) 3a + + c (a + 3) 3b Let 0 = a 0 < a < < a n < a n+ = be such that a + a + + a n =. Prove that a a a 0 + a a 3 a + + a n a n+ a n a n. 3 RESONANCE December 06

15 4. Let x, x,...,x n be positive real numbers with x + x + + x n =. Then show that n i= x i x i n n. 5. If a, b, c (0, ) satisfy a + b + c =, prove that abc ( a)( b)( c) Let a, b, c be positive real numbers. Prove that a a + b + c + b b + c + a + c c + a + b a b + c + b c + a + c a + b. 7. Let a, b, c be positive real numbers such that ab + bc + ca = 3. Prove that ( (3a a 3 + b 3 ) + ) a + ab + ( b (3c + ) ( + (3b + ) ) c 3 + a 3 c + ca + a b 3 + c 3 b + bc + c 0abc. ) + 8. Let a, b, c be positive real numbers such that abc =. Prove that ( a + b) 4 a + b + ( b + c) 4 b + c + ( c + a) 4 c + a Let h a, h b and h c be the altitudes and r the inradius of a triangle. Prove that h a r h a + r + h b r h b + r + h c r h c + r Let a, b, c be positive numbers and abc = 8. Prove that a 4 + b 4 c 3 + b4 + c 4 a 3 + c4 + a 4 b 3 64 ( a 5 + b 5 + ) c Suggested Reading [] B J Venkatachala, Inequalities - An Approach Through Problems, Hindustan Book Agency. RESONANCE December 06 33