很不错！我来证另一边

不妨设 a>=b>=c, 设p=ab, 则p>=4
  (1/sqrt(1+a) + 1/sqrt(1+b))^2
 = [(sqrt(1+a)+sqrt(1+b))/(sqrt(1+a)*sqrt(1+b))]^2
 = ((2+a+b+2sqrt(1+a+b+ab)) / (1+a+b+ab)
 = (1+a+b+ab + 1-ab + 2sqrt(1+a+b+ab)) / (1+a+b+ab)
 = 1 + (1-ab)/(1+a+b+ab) + 2/sqrt(1+a+b+ab)
    * 设 t = 1/sqrt(1+a+b+ab)
 = 1 + (1-ab)*t^2 + 2t
 = (1-p)*t^2 + 2t + 1
 = -(p-1)*(t-1/(p-1))^2  + p/(p-1)
 <= p/(p-1)
 因此 1/sqrt(1+a) + 1/sqrt(1+b) <= sqrt(p/(p-1))
 故 m = 1/sqrt(1+a) + 1/sqrt(1+b) + 1/sqrt(1+c)
     <= sqrt(p/(p-1)) + 1/sqrt(1+c)
     = sqrt(1/(1-1/p)) + 1/sqrt(1+8/p)
 再令w=1/p,  则  0<w<=1/4
  m <= sqrt(1/1-w) + sqrt(1/(1+8w))
  m^2 <= (sqrt(1/1-w) + sqrt(1/(1+8w))^2
      <= 2(1/(1-w) + 1/(1+8w))
      = 2(2+14w-16w^2-7w+16w^2)/(1+7w-8w^2)
      = 2*2 + 2（16w^2-7w)/(1+7w-8w^2)
      = 4 + 2*16w(w-7/16)/((1+8w)(1-w))
      < 4    //注意 0<w<=1/4,   w-7/16<0  1-w>0
  因此  m < 2