Question

[Nguyễn Minh Tuyền]

Choa,b,c >0 Chứng minh \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

Answer 1

[Đinh Đức Hùng]

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(ab+ac+bc\right)\left(a+b+c\right)-9abc\ge0\)

\(\Leftrightarrow a^2b+a^2c+abc+abc+ab^2+b^2c+abc+ac^2+bc^2-9abc\ge0\)

\(\Leftrightarrow a^2b+a^2c+ab^2+b^2c+ac^2+bc^2-6abc\ge0\)

\(\Leftrightarrow\left(a^2b-2abc+bc^2\right)+\left(a^2c-2abc+b^2c\right)+\left(ab^2-2abc+ac^2\right)\ge0\)

\(\Leftrightarrow b\left(a-b\right)^2+c\left(a-c\right)^2+a\left(b-c\right)^2\ge0\)(luôn đúng \(\forall a;b;c>0\))

Vật bđt đã đc chứng minh

Answer 2

[HeroZombie]

Cho a,b,c>0 thì dễ thôi :v

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}\)

Khi a=b=c