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\(\hept{\begin{cases}a< b\Rightarrow2a< a+b\\c< d\Rightarrow2c< c+d\\m< n\Rightarrow2m< m+n\end{cases}}\)
\(\Rightarrow2\left(a+c+m\right)< a+b+c+d+m+n\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\left(đpcm\right)\)
\(\frac{a}{b}< \frac{c}{d}\)
\(\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+ab< bc+ab\)
\(\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\) (1)
\(\frac{a}{b}< \frac{c}{d}\)
\(\Leftrightarrow ad< bc\)
\(\Leftrightarrow ad+cd< bc+cd\)
\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)
Đặt \(S=\frac{a}{a+b+c}+\frac{b}{b+c+d}+\frac{c}{c+d+a}+\frac{d}{d+a+b}\)
Ta có: \(\frac{a}{a+b+c}< \frac{a}{a+c}\)
\(\frac{b}{b+c+d}< \frac{b}{b+d}\)
\(\frac{c}{c+d+a}< \frac{c}{a+c}\)
\(\frac{d}{d+a+b}< \frac{d}{d+b}\)
\(\Rightarrow S< \left(\frac{a}{a+c}+\frac{c}{a+c}\right)+\left(\frac{b}{b+d}+\frac{d}{d+b}\right)\)
\(\Rightarrow S< 2\left(1\right)\)
Lại có: \(\frac{a}{a+b+c}>\frac{a}{a+b+c+d}\)
\(\frac{b}{b+c+d}>\frac{b}{b+c+a+d}\)
\(\frac{c}{c+d+a}>\frac{c}{a+b+c+d}\)
\(\frac{d}{d+a+b}>\frac{d}{a+b+c+d}\)
\(\Rightarrow S>1\left(2\right)\)
Từ (1) và (2) \(\Rightarrowđpcm\)
Cho 6 số nguyên dương thỏa mãn : a<b<c<d<m<n
Chứng minh rằng: \(\frac{a+d}{a+b+c+d+m+n}<\frac{1}{3}\)
ta có
a<b<c=>3a<a+b+c
d<m<n=>3d<d+m+n
=>3a+3d<a+b+c+d+m+n
=>3a+3a/a+b+c+d+m+n<a+b+c+m+n+d/a+b+c+d+m+n
=>3(a+d)/a+b+c+d+m+n)<1
=>a+d/a+b+c+d+m+n<1/3 (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
=> a + c + m / a + b + c + d + m + n < 1/2 ( đpcm)
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 )