Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
$\dfrac{1}{ x+y} + \dfrac{1}{x+y} + \dfrac{1}{x+z} + \dfrac{1}{y+z} ≥ \dfrac{16}{3x +3y +2z}$
$\dfrac{1}{x+z} + \dfrac{1}{x+z} +\dfrac{ 1}{x+y} +\dfrac{1}{y+z} ≥\dfrac{ 16}{2x+3y +3z}$
$\dfrac{1}{z+y} + \dfrac{1}{z+y} + \dfrac{1}{x+z} +\dfrac{1}{x+y} ≥ \dfrac{16}{2x+3y+3z}$
Cộng theo vế ta được:
$4 \left({\dfrac{1}{x+y} + \dfrac{1}{y+z} +\dfrac{ 1}{z+x}}\right) ≥ 16\left ({\dfrac{1}{3x+3y+2z} +\dfrac{ 1}{3x+2y+3z} +\dfrac{ 1}{2x+3y +3z}}\right)$
Suy ra:
$\dfrac{1}{3x+3y+2z} +\dfrac{ 1}{3x+ 2y +3z} +\dfrac{ 1}{2x+3y +3z} ≤ \dfrac{4.6}{16} =\dfrac{3}{2} $ (đpcm)
Dấu "=" xảy ra khi $x=y=z=\dfrac{1}{3}$
