Question

Chứng minh rằng: Nếu a , b, c > 0 thì : \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{b+a}\ge\dfrac{3}{2}\)

Answer 1

Đặt b+c=x;c+a=y;a+b=z

Áp dụng BĐT \(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\), ta được

\(2\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge9\)

\(\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)\ge4,5\)

\(\)\(\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\ge4,5\)

\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}+1+1+1\ge4,5\)

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

Đẳng thức xảy ra khi và chỉ khi a=b=c