Question

Cho 6 số nguyên dương a<b<c<d<m<n

hãy chứng minh rằng \(\frac{a+c+m}{a+b+c+d+m+n_{ }}\)< \(\frac{1}{2}\)

Answer 1

Ta có \(\frac{1}{2}=\frac{a+c+m}{a+m+c+a+m+c}=\frac{a+c+m}{2.\left(a+c+m\right)}\)

          \(\frac{a+c+m}{a+b+c+d+m+n}=\frac{a+c+m}{a+c+m+d+m+n}\)

Vì a<b;c<d;m<n

=>a+c+m<b+d+n

=2(a+c+m)<a+c+m+b+d+n

=>\(\frac{a+c+m}{2.\left(a+c+m\right)}>\frac{a+c+m}{a+b+c+d+m+n}\)

=>\(\frac{1}{2}>\frac{a+c+m}{a+b+c+d+m+n}\)(ĐPCM)