Question:
If abc = 1 then show that a3 /{(c - a) * (b - a)} + b3 /{(a - b) * (c - b)} + c3 /{(b - c) * (a - c)} ≥ 3
Answer:
Given, a3 /{(c - a) * (b - a)} + b3 /{(a - b) * (c - b)} + c3 /{(b - c) * (a - c)}
= {a3 * (b - c)}/{(c - a) * (b - a) * (b - c)} + {b3 * (c - a)}/{(a - b) * (c - b) * (c - a)} + {c3 * (a - b)}/{(b - c) * (a - c) * (a - b)}
= -{a3 * (b - c)}/{(a - b) * (b - c) *(c - a)} - {b3 * (c - a)}/{(a - b) * (b - c) *(c - a)} - {c3 * (a - b)}/{(a - b) * (b - c) *(c - a)}
= -{a3 * (b - c) + b3 * (c - a) + c3 * (a - b)}/{(a - b) * (b - c) *(c - a)}
= -(a3 b - a3 c + b3 c - b3 a + c3 a - c3 b)/{(a - b) * (b - c) *(c - a)}
= {(a - b)*(b - c)*(c - a)*(a + b + c)}/{(a - b) * (b - c) *(c - a)}
= a + b + 3
≥ 3
So, a3 /{(c - a) * (b - a)} + b3 /{(a - b) * (c - b)} + c3 /{(b - c) * (a - c)} ≥ 3
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