由已知3x+4y≥4,所以3x/4+y≥1,即y≥1-3x/4,代入代数式x^2+y^2+2x中 原式≥x^2+(1-3x/4)^2+2x =x^2+1+9x^2/16-3x/2+2x =x^2+x/2+9x^2/16+1 =25x^2/9+x/2+1 =(5x/3+3/10)^2+91/100,当x=0时,原式有最小值(3/10)^2+91/100=1.