Question

[Hà Hoàng Thịnh]

CMR : với a , b , c \(\in\)N :

S = \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}\ge6\)

Answer 1

[Nguyễn Võ Anh Nguyên]

\(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)

Áp dụng BĐT Cauchy:
\(\Rightarrow\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge2\sqrt{\frac{ac}{ca}}+2\sqrt{\frac{ab}{ba}}+2\sqrt{\frac{bc}{cb}}=2+2+2=6\)

Dấu"=" xảy ra khi a=b=c

Answer 2

[Đinh Đức Hùng]

Lớp 6 chưa học bđt Cauchy nha bn kia

\(S=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}\ge6\)

\(\Leftrightarrow\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}\ge6\)

\(\Leftrightarrow\frac{a}{c}+\frac{b}{c}+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}-6\ge0\)

\(\Leftrightarrow\left(\frac{a}{b}-2+\frac{b}{a}\right)+\left(\frac{a}{c}-2+\frac{c}{a}\right)+\left(\frac{c}{b}-2+\frac{b}{c}\right)\ge0\)

\(\Leftrightarrow\frac{a^2-2ab+b^2}{ab}+\frac{a^2-2ac+c^2}{ac}+\frac{c^2-2bc+b^2}{bc}\ge0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2}{ab}+\frac{\left(a-c\right)^2}{ac}+\frac{\left(c-b\right)^2}{bc}\ge0\) (luôn đúng \(\forall a;b;c\in N\))

Vậy \(S\ge6\)