CHAPTER XXVII. INEQUALITIES. 276. The statement pq is positive is expressed algebraically by pq > 0, and the statement p q is negative is similarly expressed by p-q<0. With this notation the definition of an algebraic inequality becomes We shall call either of these two forms of inequalities the reverse of the other. The equational statement p=q, if p-q=0 is the transitional form through which either of these inequalities passes into its reverse. 277. The simpler properties of inequalities are direct consequences of the foregoing definition. I. Every inequality can be expressed either in the form p> 0, or in that of its reverse p < 0; — an obvious consequence of the definition of an inequality. II. If pq and q>r, then p>r. COR. If p>q>r> s... >%, then pz. III. Any term may be transferred from one side of an inequality to the other, provided its sign be changed. For, if prq, then by definition (p+r) — q> 0, that is, p-(q-r) > 0, and therefore p>q-r. Similarly, if p>q+s, it may be shown that p-s>q. IV. The signs of all the terms of an inequality may be changed, provided the symbol of inequality be reversed. For, if m +p-r> n + q − s, then by III., that is, -n-q+s>-m-p+r, - m − p + r < − n − q + s. 278. The terms of an inequality may be combined with the terms of an equation in accordance with the following laws : V. If p>q and r = s, then p±r>q± s. For, (p±r)-(9 ± 8) = (p − q) ± (r− s) = p − q = a positive quantity. VI. If p>q and r=s, then pr > qs and p/r>q/s, if r and s be positive; pr < qs and p/r < q/s, if r and s be negative. For, since rs, and p-q is positive by hypothesis, ... pr—qs, p/r-q/s = (p − q)r, (p−q)/r; whence it follows that prqs and p/r-q/s are positive when r and s are positive, but negative when r and s are negative. COR. 1. If q be positive, and p > qr, and r > s, then p> qs. COR. 2. If the terms of an inequality be fractional, they can be made integral by the proper multiplications. Thus, from > we derive 5 × 5>4 × 6, and from we obtain (4) × (− 6) < 5 × 5. 4 5 5 6 279. The terms of two or more inequalities may be associated with one another, but under certain important limitations. VII. If Pi>1, P2> 92, P3 > 93, Pn > In ... ... then P+P + P3 + ··· + Pn > 91 + 2 + 3 + ··· + In⋅ ... ... For, since (P1— 1), (P2 — J2), ··· (P2 — In) are all positive and ... ... so is (P+ P+ ··· + Pn) — (I1 + 2 + ··· +9) also positive, and therefore ... ... P1+ P2++ Pn > Q1 + Q2 + ··· + In From p1>q1 and p2 > 2 we cannot infer p1-1> P2-Y2• For example, 8 > 7 and 5 > 2, but 8 − 5 < 7 — 2. VIII. If p1> Q1, P2 > 92, P3 > 93, Pn n For, since p1>1 and P2 P3P is positive, P1 P2 Pn> 91 P2 P3**Pn and since p2q2 and 1P3P is positive, ... Pn .. P1 P2 Pn > 91 92P3' and proceeding in this way until all the p's on the right are replaced by the corresponding q's, we have finally From P1 >91 and p2 > q1⁄2 we cannot infer p1/q1> P2/J2For example, 8>6 and 4>2, but 8/4 <6/2. COR. If p and q be positive and n = a positive integer, and if p>q, then p" > q”. IX. If p and q be positive, n = a positive integer, and pl/n, qn, denote the positive real nth roots of p, q; and if P>q, then pl/n> q1/n. For, if this be not so, then p1/” < or = ql/n; but since p1/" and q1/" are both real and positive, from p1/"< or = it follows, by the corollary of VIII, that (p1/n)"<or = (q1/")", whence p< or q, which contradicts the = hypothesis p>q. = COR. If p and q be positive and m, n = positive integers, and if p>q, then p±m/n > q±m/n; provided, if n be not 1, only the positive real nth roots be taken. 280. The following examples illustrate some of the uses of the method of inequalities: Ex. 1. The sum of any real positive quantity and its reciprocal is greater than 2. Let the quantity be a, and let a 1/2 denote its positive square root. Then (a1/2-1/a1/2)2=a2+1/a>0; Ex. 2. For what values of x is x2 4x+3>1? Completing the square for x, we have x2-4x+3= (x − 2)2 - 4 + 3. that is, for all real values of x except the value x = 2. Ex. 3. For what values of x is (2 x − 1)/(x+2) >,=, or <1? according as {(2 x − 1) − (x + 2)}/ (x + 2) > or < 0, [by III.] [by VI.] or, according as - x> or <2, if x be negative; Ex. 4. The arithmetic mean of any two positive quantities is greater than their geometric mean. If the two quantities be a and b, the arithmetic mean is 1(a + b), the geometric mean is √(ab), and we have to prove that and dividing by 2 completes the proof. This example is a particular case of the following : |