Thay 1 = a.b.c ,ta có:
a + b + c > abc/a + abc/b + 1/c
<=> a + b + c > bc + ac + 1/c
<=>(a - ac) + (b - bc) + (c - 1/c) > 0
<=>a(1 - c) + b(1 - c) + (c - 1/c) > 0
Nhân cả hai vế với c, ta có:
<=>ac(1 - c) + bc(1 - c) + (c^2 - 1) > 0
<=>ac(1 - c) + bc(1 - c) + (c - 1)(c + 1) > 0
<=>ac(1 - c) + bc(1 - c) - (1 - c)(c + 1) > 0
<=>(1 - c)(ac + cb - c - 1) > 0
<=>(1 - c)[(ac - c) + (cb - 1)] > 0
<=>(1 - c)[c(a - 1) + (cb - abc)] > 0
<=>(1 - c)[c(a - 1) - cb(a - 1)] > 0
<=>(1 - c)(a - 1)(c - cb) > 0
<=>(1 - c)(a - 1)(1 - b)c > 0
<=>(c - 1)(a - 1)(b - 1)c > 0
Chia cả hai vế với c, ta có:
(c - 1)(a - 1)(b - 1) > 0