Question

cho 0 < a,b,c ≤ 1. Cmr: \(a+b+c+\frac{1}{abc}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+abc\)

Answer 1

Lời giải:

Vì $a,b,c\in (0;1]$ nên $ab,bc,ac\in (0;1]$

Do đó: \((ab-1)(bc-1)(ca-1)\leq 0\)

\(\Leftrightarrow (ab^2c-ab-bc+1)(ca-1)\leq 0\)

\(\Leftrightarrow a^2b^2c^2-(ab^2c+a^2bc+abc^2)+ab+bc+ac-1\leq 0\)

\(\Leftrightarrow a^2b^2c^2+ab+bc+ac\leq ab^2c+a^2bc+abc^2+1\)

\(\Leftrightarrow \frac{a^2b^2c^2+ab+bc+ac}{abc}\leq \frac{ab^2c+a^2bc+abc^2+1}{abc}\)

\(\Leftrightarrow abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq a+b+c+\frac{1}{abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c=1$