Question

[Đức Huy ABC]

Chứng minh bất đẳng thức Nesbitt cho 3 số thực dương bằng các cách khác nhau có thể:

\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)

Answer 1

[Hung nguyen]

Có nhiều cách lắm. T đơn cử 1 cách nhé

\(\sum\dfrac{a}{b+c}=\sum\dfrac{a^2}{ab+bc}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)

Answer 2

[Ngọc Hiền]

\(A=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)

3+A=\(\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}+1\)

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

đặtx=a+b;y=a+c;z=b+c

=>3+A=\(\dfrac{1}{2}\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

mà (x+y+z)(\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\))\(\ge\)9

=>3+A\(\ge\dfrac{9}{2}\)

=>A\(\ge\dfrac{3}{2}\)