OrEEATIONS.] ALGEBRA 523 . . . x s y or . iy O Cor. As a particular case, x 3 + y 3 + s 3 > Sxyz. Ex. 7. Given x^ . . . x n = y", to prove that ( 1 + x " The demonstration will be perfectly general in fact, though limited in form, if we suppose the number of quantities to be 5 ; in which case,
- & - ^ 3 = y 5 -
Make the number up to 8 by introducing three ys ; (1 + x,) (1 +x.,) >(1 + V^) 2 ys ; then example 1. /. Multiplying these products together, and combining the right hand factors two and two, > {(1 + t/aSjae&ieMl + t/x > ( 1 + Ex. 8. If the sum of n fractions makes up 1, the sum of their reciprocals is greater than the square of their number. Let ar 1 + a? J +...*.-l, then, -+_+... + - X l X 2 X n But whence Ex. 9. l (example G). (example 6)<- < 2n ^ + )> a; + - Let the numerator and denominator of this fraction be designated by N" and D. 1ST may be divided into pairs of terms, at the same distance from either end, viz., 1 + x? n , x 2 + x*"~ 2 , &c., with or without a middle term, each of which (after 1 + a 2 *) is, by example 4, less than that quantity ; the middle term, rf there be one, being less than 7^+1
- . in either case N< (l+# 2n ) . . . (1.)
Again (example 6), D >n Ijxx* . . . x- n ~ l >nx* . (2.) the fraction : ~<- ( x n + D 2?i To prove the second proposition, that the fraction is greater than ($+ - it is only necessary to multi ply up and reduce the result ; thus, n+l f 1 2n Whence the proposition. Additional Examples. Ex. 1. (# + y + s) 2 < 3(,r 2 + y 9 + z ;2 ), and generally, (x + y + z) n < Z n ~ l (x n + y n + z"). (See Induction.) Ex. 2. (x + y) (y + z) ^j?. 3. (x < + i/* + z*) . Ex. 4. (a 2 + 6 2 + c 2 ) (^ 2 + y 2 + 2 2 ) > (cw; + by + rc) 2 . j^a?. 5. The arithmetic mean of the pih powers of n positive quantities is greater than the joth power of their mean, and also greater than the mean of their combina tions p together. Ex. 6. (ax + by + czf- + (ax + cy + 6z) 2 + (bx + ay + czf + (bx + cy + azf + (ex + ay + bz)" + (<:x + by + azf > 6(ab + ac + be) (xy + xz + yz . < 6(a 2 + 6 2 + s 2 )(^ + 2 / 2 + 2 2 ). / 18. Induction. It will be noted that the numerical multiplier of the second term of the powers of a + x already obtained is the same as the index. It is easy to see that this law is general. To demonstrate the fact formally we employ the method of induction. The argument may be divided into four distinct steps 1. Inference; 2. Hypothesis; 3. Comparison; 4. Conclusion. The first step, inference, is the discovery of the pro bable existence of a law. The second step, hypothesis, is the assumption that that law holds to a certain point, up to which the opponent to the argument may be presumed to admit it. The third step consists in basing on this assumption the demonstration of the law to a stage beyond what the opponent was prepared to admit. The fourth step argues that as the law starts fair, and advances beyond a point at which any opponent is prepared to admit its existence, it is necessarily true. Ex. 1. To prove that (a + x) n = a n + na n ~ 1 x + , &c. I. By multiplication we get (a + xY = a* + 4a 3 .r + , <c. II. Let it be granted that (a + x} m = a m + ma n ~ l x + , &c., where m is the extreme limit to which the opponent will admit of its truth. III. By multiplying the equals by a + x, we get (a + #) m+1 = a m+1 + ma m x + , &c., + a"x + , &c., = a m+l + (m + l)a m x + , &c., i.e., if the law be admitted true for m it is proved true for TO + 1 ; in other words, at whatever point the opponent compels us to limit our assumption, we can advance one step higher by argument. IV. Now, the law is true for 4, . it is proved true for 5 ; and being true for 5, it is proved true for 6, and so on, ad injlnitum. Ex. 2. The sum of the cubes of the natural numbers is the square of the sum of the numbers, I. Let us assume that II. If this be so, then by adding (x+ I) 3 we get III. Hence, if the law be true for any one number ar, it
is also true for x + 1.
