(a+1)(b+1)(c+1)(a+1)(b+1)(c+1)
⇒(a+1)2(b+1)2(c+1)2(⋅)⇒(a+1)2(b+1)2(c+1)2(⋅)
Ta có (a−1)2≥0(a-1)2≥0
⇒a2−2a+1≥0⇒a2-2a+1≥0
⇒a2+2a+1≥4a⇒a2+2a+1≥4a
⇒(a+1)2≥4a⇒(a+1)2≥4a
Tương tự: (b+1)2≥4b;(c+1)2≥4c(b+1)2≥4b;(c+1)2≥4c
Thay vào (⋅)(⋅) ta có:
(a+1)2(b+1)2(c+1)2≥4a.4b.4c=64abc=64((a+1)2(b+1)2(c+1)2≥4a.4b.4c=64abc=64(Vì abc=1)abc=1)
⇒(a+1)(b+1)(c+1)≥8⇒(a+1)(b+1)(c+1)≥8 (a,b,c∈N)(a,b,c∈N)
⇒Đpcm