Question

[Nguyễn Đức Trung]

Chứng minh \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge\frac{3}{2}\)

Mọi người làm hộ mình nha.

Answer 1

[Hoàng Phúc]

thiếu đề nhé, x,y,z>0 nữa

Cần CM bđt phụ sau: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (a,b,c>0)

\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\)

Theo bđt Cô-Si: \(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\)

Tương tự: \(\frac{b}{c}+\frac{c}{b}\ge2;\frac{a}{c}+\frac{c}{a}\ge2\)

\(=>\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3+2+2+2=9\)

Vậy ta đã CM đc bđt phụ

Đặt a=y+z;b=x+z;c=x+y

=>a+b+c=2x+2y+2z=2(x+y+z)

Ta có: \(2\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge9\)

\(=>\left(x+y+z\right)\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge\frac{9}{2}\)

\(=>\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\ge\frac{9}{2}\)

\(=>\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\ge\frac{9}{2}\)

\(=>\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\ge\frac{9}{2}-3=\frac{3}{2}\)

Dấu "=" xảy ra <=>x=y=z

Vậy.........................