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Bernoulli’s inequality (for integer cases) (1 + x)n ≥ 1 + nx (1) with conditions: (a) (1) is true ∀ n∈N ∪ {0} (b) (1) is true ∀ n∈2N and ∀x∈ R, x ≥ -1. and ∀x∈ R. (1 + x)n > 1 + nx The strict inequality: is true for every integer n ≥ 2 (2) and every real number x ≥ -1 with x ≠ 0. (The strict inequality is not discussed in the following.) Proof Use Mathematical Induction 1 Condition (a) Let P(n) be the proposition: (1 + x)n ≥ 1 + nx 0 (1 + x) = 1 ≥ 1 + 0x For P(0), ∀n∈N ∪ {0} and ∀x∈ R, x ≥ -1. ∴ P(0) is true. k ≥ 0, Assume P(k) is true for some k∈Z, k that is, (1 + x) ≥ 1 + kx ∀ k∈N ∪ {0} For P(k+1), (1 + x)k+1 = (1 + x)k(1 + x) ≥ (1 + kx) (1 + x) , by (3) and ∀x∈ R, and also note that since x ≥ -1. (3) x ≥ -1, the factor (x + 1) ≥ 0. 2 = 1 + (k + 1) x + kx ≥ 1 + (k + 1) x ∴ P(k+1) is also true. By the Principle of Mathematical Induction, P(n) is true ∀n∈N ∪ {0} and ∀x∈ R, x ≥ -1. Condition (b) Let P(n) be the proposition: (1 + x)n ≥ 1 + nx (1 + x)0 = 1 ≥ 1 + 0x For P(0), 2 and ∀x∈ R. ∴ P(0) is true. 2 (1 + x) = 1 + 2x + x ≥ 1 + 2x, For P(2), ∴ P(2) ∀ n∈2N since x2 ≥ 0, ∀x∈ R. is true. Assume P(k) is true for some k∈Z, that is, k ≥ 0, k (1 + x) ≥ 1 + kx ∀ k∈N ∪ {0} and ∀x∈ R, x ≥ -1. (4) For P(k + 2), (1 + x)k+2 = (1 + x)k (1 + x)2 ≥ (1 + kx)(1 + 2x), by (4) and P(2). 2 = 1 + (k + 2) x + 2kx ≥ 1 + (k + 2) x, since k > 0 and x2 ≥ 0, ∀x∈ R. ∴ P(k + 2) is also true. By the Principle of Mathematical Induction, P(n) is true ∀ n∈2N and ∀x∈ R. Condition (a) is discussed only in the following. 1 Proof Use A.M. ≥ G.M. 2 Consider the A.M. and G.M. of n positive numbers (1 + nx), 1, 1, ….,1 [with (n-1) “1”s] (1 + nx ) + 1 + 1 + .... + 1 n n + nx n ≥ n (1 + nx ).1.1....1 ≥ n (1 + nx ) (1 + x)n ≥ 1 + nx ∴ (1) Numbers should be positive before applying A.M. – G.M. theorem. Note : In the numbers used in A.M.-G.M. above, 1 > 0 and However, if 1 + nx ≥ 0, i.e. x ≥ − since it is given that x ≥ -1, or 1 + nx < 0, 1 n . x + 1 ≥ 0, L.H.S. of (1) = (1 + x)n ≥ 0 R.H.S. of (1) = 1 + nx < 0 ∴ (1 + x)n ≥ 0 > 1 + nx, which is always true. ∴ x ≥ -1 and not x ≥ − Proof 1 n can ensure (1) is correct. Use Binomial Theorem 3 ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ (1 + x ) n = 1 + nx + ⎜ ⎟ x 2 + ⎜ ⎟ x 3 + ⎜ ⎟ x 4 + .... + ⎜ ⎟ x n ≥ 1 + nx ⎝n⎠ ⎝2⎠ ⎝ 3⎠ ⎝4⎠ (a) For x > 0, (b) For x = 0, (c) For -1 < x < 0, Put y = -x, obviously (1 + x)n ≥ 1 + nx (5) is true. (The proof below is not very rigorous.) then 0 < y < 1, ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ (1 − y ) n = 1 − ny + ⎜ ⎟ y 2 − ⎜ ⎟ y 3 + ⎜ ⎟ y 4 − .... + ( −1) n ⎜ ⎟ y n ⎝n⎠ ⎝ 2⎠ ⎝ 3⎠ ⎝4⎠ (6) Put y = 0, we have: ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ 0 = (1 − 1) n = 1 − n + ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − .... + ( −1) n ⎜ ⎟ ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝n⎠ ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ ∴ ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − .... + ( −1) n ⎜ ⎟ = n − 1 > 0 ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝n⎠ Now (7) 0 < y < 1 , ∴ y2 > y3 > …> yn Therefore in (7), each term is multiplied by a factor that is smaller than the term before, n ⎛n⎞ 2 ⎛n⎞ 3 ⎛n⎞ 4 n⎛ ⎞ n ⎜ ⎟ y − ⎜ ⎟ y + ⎜ ⎟ y − .... + ( −1) ⎜ ⎟ y ≥ 0 ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠ ⎝n⎠ (8) From (6), (1 – y)n ≥ 1 – ny or n (1 + x) ≥ 1 + nx for 0 < y < 1. for -1 < x < 0. 2 Extension of Bernoulli’s inequality Given x > -1, then (a) (b) (1 + x)r ≤ 1 + rx r (1 + x) ≥ 1 + rx for 0<r<1 for r < 0 or (9) r>1 (10) Firstly we give the proof that r is a rational number first. Use A.M. ≥ G.M. Proof 4 Since r ∈ Q, (a) p r= q Let 0 < r < 1, ∴ p < q, q – p > 0. Also 1 + x > 0, (1 + x)r = (1 + x ) = (b) p q px + q q q = (1 + x ) p 1q − p ≤ =1+ p q p(1 + x ) + 1 ⋅ ( q − p ) q (G.M. ≤ A.M.) x = 1 + rx Let r > 1, (i) If 1 + rx ≤ 0, then (ii) If 1 + rx > 0, (1 + x)r > 0 ≥ 1 + rx. rx > -1. 1 Since r > 1, we have 0 < r < 1 . By (a) we get: 1r 1 (1 + rx ) ≤ 1 + rx = 1 + x r r ∴ (1 + x) ≥ 1 + rx. Let r < 0, then - r > 0. Choose a natural number n sufficiently large such that 0 < -r/n < 1 and 1 > rx/n > -1. By (a), Since 0 < (1 + x ) −r n ≤1+ r n x > 0. −1 2 r r r r r 1 ≥ 1 − ⎛⎜ x ⎞⎟ = ⎛⎜ 1 − x ⎞⎟⎛⎜ 1 + x ⎞⎟ ⇒ ⎛⎜ 1 − x ⎞⎟ ≥ 1 + x ⎝n ⎠ ⎝ n ⎠⎝ n ⎠ ⎝ n ⎠ n (11) Hence by (11), r n r (1 + x )r ≥ ⎛⎜ 1 + x ⎞⎟ ≥ 1 + n ⎛⎜ x ⎞⎟ = 1 + rx . ⎝ n ⎠ ⎝n ⎠ Again, equality holds if and only if x = 0. Note : If r is irrational, we choose an infinite sequence of rational numbers r1, r2, r3, …., such that rn tends to r as n tends to infinity. For part (a), we can extend to irrational r: (1 + x ) r = lim (1 + x ) rn ≤ lim (1 + rn x ) = 1 + rx . n→∞ n→∞ Similar argument for part (b) completes the proof for the case where r ∈ R. 3 Proof Use analysis 5 f(x) = (1 + x)r – 1 – rx Let where x > –1 and r ∈ R\ {0, 1} (12) Then f(x) is differentiable and its derivative is: f ’(x) = r(1 + x)r-1 – r = r [(1 + x)r-1 – 1] from (13) we can get (a) If 0 < r < 1, f ’(x) = 0 ⇔ x = 0. then f ’(x) > 0 ∀ x∈ (–1, 0) and ∴ x = 0 is a global maximum point of (b) ∴ f(x) < f(0) = 0. ∴ (1 + x)r ≤ 1 + rx If (13) for f. 0 < r < 1. r < 0 or r > 1, then f ’(x) < 0 ∀ x∈ (-1, 0) and ∴ x = 0 is a global minimum point of ∴ f(x) > f(0) = 0. ∴ (1 + x)r ≥ 1 + rx for f ’(x) < 0 ∀ x∈ (0, +∞). r < 0 or f ’(x) > 0 ∀ x∈ (0, +∞). f. r > 1. Finally, please check that the equality holds for x = 0 or for r ∈ {0, 1}. The proof is complete. 4

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