Q.88. Statistics, To prove that rms ≥ arithmetical mean

Question 88.
Prove that the root mean square value ≥ mean value (for any given data).

Answer 88.
For two numbers a and b
(a - b)^2 ≥ 0
=> a^2 + b^2 - 2ab ≥ 0
=> 2a^2 + 2b^2 ≥ a^2 + 2ab + b^2
=> 2 (a^2 + b^2) ≥ ( a + b )^2
=> (a^2 + b^2)/2 ≥[(a + b)/2]^2
=> √[(a^2 + b^2)/2] ≥ (a + b)/2
Thus, for two quantities a and b,
the root mean square value ≥ mean value.

Now, let us see how to prove for 3 quantities.
For 3 quantities, a, b and c,
(a - b)^2 + (b - c)^2 + (c - a)^2 ≥ 0
=> 2(a^2 + b^2 + c^2) ≥ 2 (ab + bc + ca)
=> 3(a^2 + b^2 + c^2) ≥ a^2 + b^2 + c^2 + 2 (ab + bc + ca)
=> 3(a^2 + b^2 + c^2) ≥ (a + b + c)^2
=> (a^2 + b^2 + c^2)/3 ≥ (a + b + c)^2/9
=> (a^2 + b^2 + c^2)/3 ≥ [(a + b + c)/3]^2
=> √[(a^2 + b^2 + c^2)/3] ≥ (a + b + c)/3
Thus, for three quantities a, b and c
the root mean square value ≥ mean value.

From here onwards things start getting more complicated. To explain, let us do for 4 quantities, a, b, c and d.
Now 2 out of these 4 can be selected in 4C2 = 6 ways. Taking all these 6 cominations,
(a - b)^2 + (b - c)^2 + (c - d)^2 + (d - a)^2 + (a -c )^2 + (b - d)^2 ≥ 0
=> 3(a^2 + b^2 + c^2 + d^2) ≥ 2(ab + bc + cd + da + ac + bd)
=> 4(a^2 + b^2 + c^2 + d^2) ≥ a^2 + b^2 + c^2 + d^2 + 2(ab + bc + cd + da + ac + bd)
=> 4(a^2 + b^2 + c^2 + d^2) ≥ (a + b + c + d)^2
=> (a^2 + b^2 + c^2 + d^2)/4 ≥ [(a + b + c + d)/4]^2
=> √[(a^2 + b^2 + c^2 + d^2)/4] ≥ [(a + b + c + d)/4.

Thus, when you have to prove for n quantities, you have to take all possible combinations of 2 out of n quantities, say, a(p) and a(q), then
Σ [ a(p) - a(q) ]^2 ≥ 0, [ p from 1 to n and q from 1 to n, p < q ]
=> (n - 1) Σ [a(p)]^2, (p = 1 to n) ≥ 2 Σ [a(p)a(q)], [ p from 1 to n and q from 1 to n, p < q ]
=> n Σ [a(p)]^2, (p = 1 to n) ≥ Σ [a(p)]^2, (p = 1 to n) + 2 Σ [a(p)a(q)], [ p from 1 to n and q from 1 to n, p < q ]
=> Σ [[a(p)]^2]/n, (p = 1 to n) ≥ [Σ [a(p)]/n]^2, ( p = 1 to n)
Thus, for 'n' quantities,
the root mean square value ≥ mean value

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