Let a1, a2, …, an be n positive real numbers (not all equal) and let m be a real number. Then a1m+a2m+...+anm/n > (a1+a2+...+an/n)m if m ? R – [0, 1].
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.
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