The Cauchy-Schwarz inequality

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1 The Cauchy-Schwarz inequality Finbarr Holland, April, Introduction The inequality known as the Cauchy-Schwarz inequality, CS for short, is probably the most useful of all inequalities, and arises in most areas of Mathematics. It comes in various shapes and sizes, and can involve finite or infinitely many variables, and be discrete or continuous, and a mixture of all of these kinds. Here we discuss the simplest kind, which involves two vectors in the same finite-dimensional vector space. The statement Let a = (a 1, a,..., a n ), b = (b 1, b,..., b n ) be two vectors with real coordinates, i.e., a, b R n. Their inner-product is the expression < a, b >= a 1 b 1 + a b + + a n b n = a i b i, and their lengths are given by a = < a, a > = n a i, b = < b, b > = n b i. A concise version of CS reads as follows:- Theorem 1. For all vectors a, b R n, < a, b > a b, with equality iff there are real numbers λ, µ, not both equal to zero, such that λa = µb. 1

2 There are many proofs of this result. We outline two. First Proof. We present one that relies on three basic facts. One is that the square of every real number is nonnegative. Another is that the sum of nonnegative numbers is nonnegative. The third is that a real nonnegative quadratic polynomial has either no real root or two coincident real roots, and the condition that one of these options occurs is that the discriminant of the quadratic be nonpositive. First of all, note that if a b = 0, then one of a, b is the zero vector, in which case their inner-product is also zero, and the inequality becomes an equality. So, it s enough to establish the result when a b > 0. To do so, let t be any real number, and note that then 0 (ta i b i ) p(t). Now, for all real t, (ta i b i ) = (t a i ta i b i + b i ) = t a i t a i b i + b i ) = t a t < a, b > + b. Thus, p is the quadratic polynomial whose values are nonnegative: p(t) = t a t < a, b > + b 0, t R. This means that, since its graph lies in the upper-half plane, either p has no real root or precisely one. In other words, its discriminant ( < a, b >) 4 a b is nonpositive. It follows that < a, b > = < a, b > a b = a b = a b. The inequality is strict unless the discriminant of p is zero. If this occurs, then p has only one real root, viz., < a, b > / a, in which case 0 = ( < a, b > a a i b i ), which forces < a, b > a i a b i = 0, i = 1,,..., n, i.e., λa = µb, where λ =< a, b >, µ = a. Second Proof. This relies on the observation that, if t > 0, ab ta + b, a, b R, (1) t

3 with equality iff ta = b. As before, we may assume that a b > 0. Applying (1), it follows that a i b i ta i + b i, i = 1,,..., n, t whence, adding these we infer that, for all t > 0, But, for all t > 0, < a, b > t a + b. () t a b t a + b, t with equality iff t = b / a. Since the left-hand side of () is independent of t it follows that < a, b > a b, < a, b > a b. Replacing a in this by its additive inverse a = ( a 1, a,..., a n ), and noting that a = a we deduce that so that < a, b >=< a, b > a b = a b, a b < a, b > a b, < a, b > a b. This establishes the inequality. If the equality sign prevails, then either < a, b >= a b or < a, b >= a b. If the former happens, there is equality in () when t a = b, and so in each of the preceding inequalities for this same t. In other words, only when a a i = b b i, i = 1,,..., n. Example 1. Suppose a, b are real numbers. Then, for all real x, a cos x + b sin x a + b, with equality iff a sin x = b cos x. Solution. Apply the CS inequality with (a 1, a ) = (a, b), (b 1, b ) = (cos x, sin x). Then a cos x + b sin x a + b cos x + sin x = a + b, since cos x + sin x = 1, x R.

4 Equality can occur iff there are constants λ, µ, not both equal to zero, such that λa = µ cos x, λb = µ sin x, i.e., a sin x = b cos x, i.e., tan x = b, if a 0. a Example. Suppose a, b, c are real numbers. Then, for all real x, y, a cos x cos y + b sin x cos y + c sin y a + b + c. Solution. Apply the CS inequality with Then (a 1, a, a ) = (a, b, c), (b 1, b, b ) = (cos x cos y, sin x cos y, sin y). a cos x cos y + b sin x cos y + c sin y a + b + c cos x cos y + sin x cos y + sin y = a + b + c (cos x + sin x) cos y + sin y = a + b + c (cos y + sin y = a + b + c, with equality, for some pair of real numbers x, y, iff there are constants λ, µ, not both equal to zero, such that λa = µ cos x cos y, λb = µ sin x cos y, λc = µ sin y. Exercise 1. Let r > 0. Show that, for all real x, y, the point (r cos x cos y, r sin x cos y, r sin y) lies on the sphere centred at (0, 0, 0) whose radius is r. Example. Let a, b, c, d be real numbers with a + b + c > 0. Determine the distance from the origin to the plane Answer: the distance is equal to M = {(x, y, z) R : ax + by + cz = d}. d a + b + c. For, if P = (x, y, z) M, then, by two applications of Pythagoras theorem, its distance from the origin is equal to x + y + z. But ax + by + cz = d, 4

5 and, by the CS inequality, d = ax + by + cz a + b + c x + y + z. Hence x + y + z d a + b + c. This means that no point in M lies inside any sphere centred at (0, 0, 0) whose radius is less than d / a + b + c. In addition, there is equality in the CS inequality, for some point P M, iff x + y + z = d, and ax + by + cz = d. a + b + c In other words, iff there are constants λ, µ, not both of which are zero, such that λa = µx, λb = µy, λc = µz, and ax + by + cz = d. In geometrical language, iff P is a point of intersection of the sphere centred at (0, 0, 0) whose radius is equal to d / a + b + c and the plane M, equivalently, iff P is the point of tangency of the plane with the sphere. Corollary 1. Let a, b, c, d be real numbers with a +b +c > 0. Let P 0 = (x 0, y 0, z 0 ) be any point in R. Show that the distance from P 0 to the plane M = {(x, y, z) R : ax + by + cz = d} is given by ax 0 + by 0 + cz 0 d a + b + c. Remark. This is the analogue in -space, R, of a result in the plane, R, with which you are familiar from school. Example 4. Suppose, in the usual notation, that a, b, c are the lengths of the sides of a triangle ABC. Let m a, m b, m c denote the lengths of the medians from the vertices A, B, C, respectively, to the opposite sides. Then am a + bm b + cm c with equality iff ABC is equilateral. (a + b + c ),.Solution. We begin by establishing a formula for the length of the median from A. Let D be the mid-point of the side BC. Since BD = a = DC, by two applications of the Cosine Rule, m a + ( a ) c am a = cos ADB = cos ADC = m a + ( a ) b am a. 5

6 Hence Similarly, by cyclicity, m a = b + c a. and so, in particular, Hence, by the CS inequality, m b = c + a b m c = a + b c, am a + bm b + cm c m a + m b + m c = 4 (a + b + c ). = ma + m b + c m a + b + c (a + b + c ), with equality iff there are constants λ, µ, not both of which are zero, such that λm a = µa, λm b = µb, λm c = µc. But neither λ nor µ can be zero. Hence there is equality iff am a + bm b + cm c = and there is a constant ν > 0 such that (a + b + c ), m a = νa, m b = νb, m c = νc. It follows that ν = /, whence m a = a, m a = 4 a, b + c a = a 4, i.e., a = b + c. Similarly, b + c + a and c = a + b. From these equations it follows that a = b = c, as desired. Exercise. In the same notation, show that m a m b = 9 a b, m 4 a = 9 a Exercise. In the same notation, show that a, b, c and m a, m b, m c are oppositely ordered Exercise 4. In the same notation, show that m a, m b, m c are the lengths of the sides of a triangle, whose area is equal to 4. 6

7 Exercise 5. In the same notation, show that with equality iff a = b = c. am a + bm b + cm c 1 (a + b + c)(m a + m b + m c ), Example 5. Let P be an arbitrary point within a triangle ABC whose area is. Denote the distances from P to the sides BC, CA and AB by x, y, z, respectively. Then x + y + z 4 a + b + c, with equality iff P is such that x a = y b = z c = a + b + c. Solution. The triangle ABC is the union of the triangles BP C, CP A, AP B, which are disjoint if we ignore common sides, which have zero area. Hence the area of ABC is the sum of the areas of BP C, CP A, AP B. Therefore i.e., so that with equality iff and for some ν. This happens iff = 1 xa + 1 by + 1 zc, = ax + by + cz a + b + c x + y + z, x + y + z x + y + z = 4 a + b + c, 4 a + b + c, x = νa, y = νb, z = νc ν = a + b + c. Hence the point P which furnishes the minimum of x + y + z is such that the lengths of the perpendiculars from it to the sides of the triangle are proportional to the lengths of the sides: x a = y b = z c = a + b + c. This point is known as the Lemoine point of the triangle. 7

8 Definition 1. In the same notation: Let x be the distance from the Lemoine point to BC. The angle ω such that is called the Brocard angle of ABC. Exercise 6. Prove that tan ω = x a, cot ω = cot A + cot B + cot C 8

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