Question

a, cho x,y,z là các số dương.

c/m: \(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+z+x}+\dfrac{z}{2z+x+y}\le\dfrac{3}{4}\)

b, cho a,b,c thỏa mãn: a+b+c=0. c/m: ab+bc+ca\(\le\)0

Answer 1

Câu b mình vừa làm rồi

a)

Áp dụng bđt Cauchy-Scharz:

\(\dfrac{x}{2x+y+z}+\dfrac{y}{2y+x+z}+\dfrac{z}{2z+x+y}\)

\(=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(x+y\right)+\left(y+z\right)}+\dfrac{z}{\left(x+z\right)+\left(y+z\right)}\)

\(\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\)

\(=\dfrac{1}{4}.3=\dfrac{3}{4}\)

Dấu "=" khi \(x=y=z\)