LI. INEQUALITIES.

667. It is often useful to know which is the greater of two given expressions; propositions relating to such questions are usually collected under the head Inequalities.

We

say that α is greater than b when a-b is a positive

quantity. See Art. 95.

668. An inequality will still hold after the same quantity has been added to each member or taken from each member.

> b,

For suppose a>

therefore a b is positive,

therefore ac- (b+c) is positive,

therefore a±c>b±c.

Hence we may infer that a term may be removed from one member of an inequality and affixed to the other with its sign changed.

669. If the signs of all the terms of an inequality be changed the sign of inequality must be reversed.

For to change all the signs is equivalent to removing each term of the first member to the second, and each term of the second member to the first.

670. An inequality will still hold after each member has been multiplied or divided by the same positive quantity.

For suppose a >b, therefore a -b is positive, therefore if m be positive m (a-6) is positive, therefore ma> mb; and similarly (ab) is positive, and > •

1

m

α

b

m m

In like manner we can shew that if each member of an inequality be multiplied or divided by the same negative quantity, the sign of inequality must be reversed.

671. If ab, a'>b', a">b", ...... then

"

a + a + a′′ +

>b+b'+b′′ +

......

For by supposition, a-b, a-b, a"-b",...... are all positive; therefore a- b + a' — b' + a′′ – b′′ +

......

is positive; therefore

672. If a>b, a'> b′, a">b", ...... and all the quantities are positive, then it is obvious that aa'a"......> bb'b"

......

673. If a>b, and a and b are positive, then a">b", where n is any positive quantity.

This follows from the preceding Article if n be an integer. If n

and we have to prove that h1>k; this we can prove indirectly;

for if h; then hk, and if hk, then h<k; both of

these results are false; hence we must have h1 > k1.

If n be a negative quantity, let n=-m, so that m is positive;

nominators are all of the same sign, then the fraction

lies in magnitude between the least and greatest of the fractions

magnitude, and suppose that all the denominators are positive;

then

In like manner the theorem may be established when all the denominators are supposed negative.

If

1

=

b2

=

аз

b

3

then each of these fractions is equal to

the fraction whose numerator is the sum of the numerators and denominator the sum of the denominators.

675. Since (xy) or x2 - 2xy + y2 is a positive quantity or zero, according as x and y are unequal or equal, we have

§ (x2 + y3) > xy,

the inequality becoming an equality when x=y. Hence

1⁄2 (a + b) > √(ab);

that is, the arithmetic mean of two quantities is greater than the geometric mean, the inequality becoming an equality when the two quantities are equal.

676. Let there be n positive quantities, a, b, c, ... k; then

unless the n quantities are all equal, and then the inequality becomes an equality.

By proceeding in this way we can shew that if p be tive integral power of 2,

abcd... (p factors)<

Now let p=n+r, and let

p

any posi

suppose each of the remaining r quantities out of the p quantities

to be equal to t; we have then

<(nt +re)'

abcd... (n factors) × ť <

n+r rt

; that is, <t+*;

(a + b + c + d + ...)".

therefore abcd... (n factors) <t"; that is, <

n

Thus the theorem is proved whatever be the number of quantities a, b, c, d, ... The inequality becomes an equality when all the n quantities are equal.

We may also write the theorem thus,

by extending the signification of the terms arithmetical mean and geometrical mean, we may enunciate the theorem thus; the arithmetical mean of any number of positive quantities is greater than the geometrical mean.

677. The following proof of the theorem given in the preceding Article will be found an instructive exercise.

Suppose a and b respectively the greatest and least of the n quantities a, b, c, d,.........k; let a1 = b1 = (a+b), and let

2

1

1

P1 = (a,b,cd......k); then since a,b,> ab, we have P,> P. Next if the factors in P, be not all equal, remove the greatest and least of them, and put in their places two new factors, each equal to half the sum of those removed; let P, denote the new geometrical mean; then P1>P1. If we proceed in this way, we obtain a series P, P1, P, P, ...... P., each term of which is greater than P ̧‚..............P., the preceding term; and by taking r large enough, we may have the factors which occur in P, as nearly equal as we please; thus when r is large enough, we may consider P=Q; therefore P is less than Q.

678. We will now compare the quantities

We suppose a and b positive, and a not less than b.

564), so that we have a result which is arithmetically intelligible and true. Hence if m be negative or any positive integer, it follows