14 1. Preliminary Notions Proposition 1.16. (AGM inequality) Given positive numbers x1,...,xr, (1.18) r i=1 xi 1/r ≤ 1 r r i=1 xi , where equality holds if and only if x1 = x2 = · · · = xr. Proof. We give here a proof due to P´ olya. It is based on the inequality ex ≥ 1 + x which is easily shown to be valid for all x ∈ R, with equality if and only if x = 0. Given x1,...,xr ∈ R+, let M and P stand for their arithmetic and geometric mean, respectively. If x1 = x2 = . . . = xr, then M = P and we are done with the last claim. Therefore, it remains to prove that P M if we assume that the xi’s are not all equal. Note that M 0. In the inequality ex ≥ 1 + x, replace x by xi/M − 1, so that (1.19) exp xi M − 1 ≥ xi M for i = 1, 2,...,r. Now, since the xi’s are not all equal, it follows that xj M for at least one j ∈ [1,r], implying that xj/M − 1 0. This means that at least one of the inequalities in (1.19) is strict. Therefore, multiplying all the inequalities in (1.19), we get r i=1 exp xi M − 1 r i=1 xi M , exp 1 M r i=1 xi − r i=1 1 1 M r r i=1 xi , er−r P r M r , implying that M P , which is what we wanted to prove. Proposition 1.17. (Cauchy-Schwarz inequality) Given two sets of real num- bers a1,...,ak and b1,...,bk, (1.20) k i=1 aibi 2 ≤ k i=1 a2 i k i=1 b2 i . Proof. This result can be proved by noticing that the quadratic polynomial t2 k i=1 a2 i − 2t k i=1 aibi + k i=1 b2 i = k i=1 (ait − bi)2

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