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in Linear Inequations by (23.6k points)
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If a, b, c, d are four distinct positive real numbers and if 3s = a + b + c + d, then

(a) abcd > 81 (s – a) (s – b) (s – c) (s – d)

(b) abcd < 9 (s – a) (s – b) (s – c) (s – d)

(c) abcd < 18 (s – a) (s – b) (s – c) (s – d)

(d) abcd < 27 (s – a) (s – b) (s – c) (s – d)

1 Answer

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Best answer

(a) abcd > 81(s – a) (s – b) (s – c) (s – d)

3s = a + b + c + d ⇒ 3s – b – c – d = a 

⇒ a = (s – b) + (s – c) + (s – d) 

For distinct positive real numbers AM > GM

⇒ \(\frac13\)[(s – b) + (s – c) + (s – d)]> {(s – b) + (s – c) + (s – d)}\(\frac13\)

⇒ (s – b) + (s – c) + (s – d) > 3 {(s – b) + (s – c) + (s – d)}\(\frac13\)

⇒ a > 3 {(s – b) + (s – c) + (s – d)}\(\frac13\)               ....(i)

Similarly,  b > 3 {(s – a) + (s – c) + (s – d)}\(\frac13\)               ....(ii)

 c > 3 {(s – a) + (s – b) + (s – d)}\(\frac13\)               ....(iii)

 a > 3 {(s – a) + (s – b) + (s – c)}\(\frac13\)               ....(iv)

∴ (i) × (ii) × (iii) × (iv)

⇒ abcd > 81{(s - a)3 (s – b)3 (s – c)3 (s – d)3}\(\frac13\)

abcd > 81(s – a) (s – b) (s – c) (s – d).

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