Q.243. Algebraic inequality.

Question 243.
If a, b, c, d are distinct positive real numbers such that 3s = a + b + c + d, then prove that abcd > 81(s - a)(s - b)(s - c)(s - d).

Answer 243.
3s = a + b + c + d
=> a = 3s - b - c - d = (s - b) + (s - c) + (s - d)
 Now, AM ≥ GM
=> (1/3) [ (s - b) + (s - c) + (s - d) ] ≥ [ (s - b)(s - c)(s - d) ]^(1/3)
=> a ≥ 3 [ (s - b)(s - c)(s - d) ]^(1/3) ... (1)

Similarly,
b ≥ 3 [ (s - c)(s - d)(s - a) ]^(1/3) ... (2)
c ≥ 3 [ (s - d)(s - a)(s - b) ]^(1/3) ... (3)
d ≥ 3 [ (s - a)(s - b)(s - c) ]^(1/3) ... (4)

Multiplying equations (1), (2), (3) and (4),

abcd ≥ 81 [ (s - a)^3(s - b)^3(s - c)^3(s - d)^3 ]^(1/3)
=> abcd ≥ 81(s - a)(s - b)(s - c)(s - d)

As a, b, c and d are distinct positive real numbers, s - a, s - b, s - c and s - d are also distict positive numbers. Hence, only inequality holds.
=> abcd > 81 (s - a)(s - b)(s - c)(s -d).

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