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Do a < b < c < d < m < n
=> a + c + m < b + d + n
=> 2 × (a + c + m) < a + b + c + d + m + n
=> a + c + m / a + b + c + d + m + n < 1/2 ( đpcm)
Do a < b < c < d < m < n
=> a + c + m < b + d + n
=> 2 × (a + c + m) < a + b + c + d + m + n
=> a + c + m / a + b + c + d + m + n < 1/2 ( đpcm)
Do a<b<c<d<m<n
=>a+c+m<b+d+n
=>2(a+c+m)<a+b+c+d+m+n
=>\(\frac{2\left(a+c+m\right)}{a+b+c+d+m+n}
a<b=>2a<a+b
c<d=>2c<c+d
m<n=>2m<m+n
=>2(a+c+m)<a+b+c+d+m+n
=>\(\frac{2\left(a+c+m\right)}{a+b+c+d+m+n}
\(\hept{\begin{cases}a< b\Rightarrow2a< a+b\\c< d\Rightarrow2c< c+d\\m< n\Rightarrow2m< m+n\end{cases}}\)
\(\Rightarrow2\left(a+c+m\right)< a+b+c+d+m+n\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\left(đpcm\right)\)
Cho 6 số nguyên dương a < b < c < d < m < n
Chứng minh rằng \(\frac{a+c+m}{a+b+c+d+m+n}<\frac{1}{2}\)
a < b \(\Rightarrow\) 2a < a + b ; c < d \(\Rightarrow\) 2c < c + d ; m < n \(\Rightarrow\) 2m < m + n
Suy ra 2a + 2c + 2m = 2(a + c + m) < (a + b + c + d + m + n). Do đó
\(\frac{a+c+m}{a+b+c+d+m+n}
a < b \(\Rightarrow\) 2a < a + b
b < d \(\Rightarrow\) 2b < c + d
m < n \(\Rightarrow\) 2m < m + n
\(\Rightarrow\) 2a + 2b + 2m = 2 ( a + b + m ) < ( a + b + c + d + m + n ) . Do đó
a + b + m/a + b + c + d + m + n < 1/2 \(\Rightarrow\) ( đpcm )
a < b < c < d < m
=> a + d < c + m + n
=> 3 ( a + d ) < a + b + c + d + m + n
\(\Rightarrow\frac{3\left(a+d\right)}{a+b+c+d+m+n}< 1\)
\(\Rightarrow\frac{a+d}{a+b+c+d+m+n}< \frac{1}{3}\) ( Đpcm )
a < b => 2a < a + b ; c < d => 2c < c +d ; m < n =>2m < m + n
Suy ra 2a + 2c + 2m = 2.(a+c+m) < a + b + c + d + m + n. Do đó :
\(\frac{a+c+m}{a+b+c+d+m+n}
Ta có : a<b => a+a < a+b
=> 2a < a+b (1)
c<d => c+c < c+d
=> 2c < c+d (2)
m<n => m+m < m+n
=> 2m < m+n (3)
Từ (1); (2) và (3). => 2a + 2c +2m < a+b+c+d+m+n
=> 2(a+c+m) < a+b+c+d+m+n
=> \(\frac{a+c+m}{a+b+c+d+m+n}\)< \(\frac{1}{2}\)( đpcm)
Vì a<b;c<d;m<n
=>a+c+m<b+d+n
=>a+a+c+c+m+m<a+b+c+d+m+n
=>2a+2c+2m<a+b+c+d+m+n
=>2(a+c+m)<a+b+c+d+m+n
=>\(\frac{a+c+m}{2\left(a+c+m\right)}>\frac{a+c+m}{a+b+c+d+m+n}\)
=>\(\frac{a+c+m}{a+b+c+d+m+n}