Question

Chứng minh bất đẳng thức với a, b, c là số dương

a, (a+b+c) \((\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})\)\(\ge9\)

b, \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge1,5\)

Answer 1

\(a,\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)\(\Leftrightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge3+2+2+2=9\Rightarrowđpcm\)b, Đặt \(x=b+c;y=a+c;a+b=z\)

Khi đó :

\(=\dfrac{1}{2}\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)-3\right]\) \(\ge\dfrac{1}{2}\left(2+2+2-3\right)=1,5\Rightarrowđpcm\)

Answer 2

Cách 2:

a, Áp dụng bất đẳng thức\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) ta có:

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right)\dfrac{9}{a+b+c}=9\)

Dấu " = " khi a = b = c = 1

\(\Rightarrowđpcm\)

b, \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)

Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) ta có:

\(\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)\ge\left(a+b+c\right)\dfrac{9}{2\left(a+b+c\right)}\)

\(=\dfrac{9}{2}-3=4,5-3=1,5\)

Dấu " = " khi a = b = c

\(\Rightarrowđpcm\)

Answer 3

Cách 2:
a, Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) ta có:

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{a+b+c}=9\)

Dấu " = " khi a = b = c = 1

\(\Rightarrowđpcm\)

b, \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}+\dfrac{a+b+c}{a+b}-3\)

\(=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\)

Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) ta có:

\(\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right)-3\ge\left(a+b+c\right).\dfrac{9}{2\left(a+b+c\right)}-3=\dfrac{9}{2}-3=1,5\)

Dấu " = " khi a = b = c

\(\Rightarrowđpcm\)