The concept of arithmetic mean has been already discussed in the previous sections. This concept can be extended till the mth power for terms and such mean is termed as the arithmetic mean of mth power. Mathematically, the arithmetic mean for mth power is defined as
If a1, a2, …, an be n positive real numbers (not all equal) and let m be a real number.
However if m ∈ (0, 1), then a1m + a2m+...+ anm/n < (a1+a2+...+an/n)m.
Obviously if m ∈ {0, 1}, then a1m + a2m+...+anm/n = (a1+a2+...+an/n)m.
Illustration:
If a, b, c are positive real numbers such that a + b + c = 1, then prove that a/b + c + b/c+a + c/a+b > 3/2.
Solution: We have to show that (a/b+c+1)+(b/c+a + 1)+(c/a+b + 1) > 3/2 + 3.
i.e. 1/b+c + 1/c+1 + 1/a+b > 9/2.
Now A.M. of mth power > mth power arithmetic mean (m = – 1 here)
⇒ 1/3 [(b+c)–1 + (c+a)–1 + (a+b)–1] > [(b+c) + (c+a) + (a+b)/3]–1
⇒ 1/b+c + 1/c+a + 1/a+b > 3.3/2(a+b+c) = 9/2.
Illustration: Prove that
1m + 2m + 3m + ….. + nm > n[1/2(n+1)]m.
Solution: Taking ‘n’ quantities into consideration we have
Then as stated above, arithmetic mean of mth powers > mth power of the arithmetic mean
(1m + 2m + 3m + ….. + nm) / n > [(1+2+3+…. n) / n]m
or (1m + 2m + 3m + ….. + nm) > n [ {n(n+1)/2}/n]m
{Since the sum of first n natural numbers is [n(n+1)/2]}
Hence, (1m + 2m + 3m + ….. + nm) > n [ (n+1)/2}]m.
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You may also like to refer Geometric Mean and Harmonic Mean.
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