Question

cho a,b,c,d >0 . cmr: 

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

Answer 1

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\)

\(\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\). Mà theo BĐT AM-GM ta có:

\(\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}=\frac{\left(a+b+c+d\right)^2}{2\left[\left(a+b\right)\left(c+d\right)+\left(a+c\right)\left(b+d\right)+\left(a+d\right)\left(b+c\right)\right]}\ge\frac{2}{3}\)

Đẳng thức xảy ra khi a=b=c=d

 

Answer 2

là \(\frac{2}{3}\) nha