If a, b, c are positive real numbers, then prove that
(i) b 2 c 2 + c 2 a 2 + a 2 b 2 ≥ abc (a + b + c).
(ii) 
+ 
+ 
≥ 
(iii) 
+ 
+ 
≥ 
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(i) b 2 c 2 + c 2 a 2 + a 2 b 2 ≥ abc (a + b + c) ⇒ (bc – ca) 2 + (ab – bc) 2 + (ca – ab) 2 ≥ 0
∴ b 2 c 2 + c 2 a 2 + a 2 b 2 ≥ abc (a + b + c)
Sol. If a, b, x, y are real number and x, y > 0, then 
+ 
≥ 
.........(i)
as on solving it we have (ay – bx) 2 ≥ 0
Similarly, we can extend the inequality to three pairs of numbers
i.e. 
+ 
+ 
≥ 
Now we use this result to following questions
(ii) As 
+ 
+ 
= 
+ 
+ 
≥ 
Now, we have to prove that 
≥ 
or a 2 + b 2 + c 2 ≥ ab + bc + ca ⇒ (a – b) 2 + (b – c) 2 + (c – a) 2 ≥ 0
(iii) L.H.S. = 
+ 
+ 
≥ 
L.H.S. = 
+ 
+ 
≥ 
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