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\(\frac{a}{b}< \frac{c}{d}\)
\(\Rightarrow ad>bc\)
\(\Rightarrow ad+ab< bc+ab\)
\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\) (1)
\(\Rightarrow ad+cd< bc+cd\)
\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\) (2)
Từ (1); (2) => \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)
\(\frac{a}{b}< \frac{c}{d}\)
\(\Rightarrow ad=bc\)
\(\Rightarrow ad+ab< bc+ab\)
\(\Rightarrow a\left(b-d\right)< b\left(a+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)
\(\Rightarrow ad+cd< bc+cd\)
\(\Leftrightarrow d\left(a+c\right)< c\left(b+d\right)\)
\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(2\right)\)
Từ ( 1 ) và ( 2 )
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)( đpcm )
Có \(\frac{a}{b}< \frac{c}{d}=>a.d< c.b\)
<=>2018a.d<2018c.b
<=>2018a.d+c.d<2018c.b+c.d
<=>d(2018a+c)<c(2018b+d)
<=>đpcm
Ta có:a/b<c/d<=>a.d<b.c
<=>2018a.d<2018b.c
<=>2018a.d+c.d<2018b.c+d.c
<=>d(2018a+c)<c(2018b+d)
<=>2018a+c/2018b+d<c/d(dpcm)
Ta có: Để \(\frac{2018\cdot a+c}{2018\cdot b+d}< \frac{c}{d}\Rightarrow\left(2018\cdot a+c\right)\cdot d< \left(2018\cdot b+d\right)\cdot c\)
\(2018\cdot a\cdot d+c\cdot d< 2018\cdot b\cdot c+c\cdot d\)
\(2018\cdot a\cdot d< 2018\cdot b\cdot c\)(bỏ cả 2 vế đi \(c\cdot d\))(gọi là (1))
Vì \(\frac{a}{b}< \frac{c}{d}\Rightarrow a\cdot d< b\cdot c\Rightarrow2018\cdot a\cdot d< 2018\cdot b\cdot c=\left(1\right)\)Mà (1) bằng \(\frac{2018\cdot a+c}{2018\cdot b+d}< \frac{c}{d}\) (điều phải chứng minh)
Do a < b < c < d < m < n
=> 2c < c + d
m< n => 2m < m+ n
=> 2c + 2a +2m = 2 ( a + c + m) < a +b + c + d + m + n)
Do đó :
(a + c + m)/(a + b + c + d + m + n) < 1/2(đcpcm)
Từ:\(\hept{\begin{cases}a< c\\c< d\\m< n\end{cases}}\Rightarrow a+c+m< c+d+n\)
\(\Rightarrow2\left(a+c+n\right)< a+b+c+d+m+n\)
\(\Rightarrow\frac{a+c+m}{a+b+c+d+m+n}< \frac{1}{2}\)
Ta có:
\(\frac{a}{b+c+d}>\frac{a}{a+b+c+d};\frac{b}{a+c+d}>\frac{b}{a+c+b+d};\frac{c}{b+c+d}>\frac{c}{a+b+c+d}\)
\(\frac{d}{a+b+c}>\frac{d}{a+b+c+d}\)
\(\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+c+b+d}\)
\(\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}>\frac{a+b+c+d}{a+b+c+d}=1\left(1\right)\)
Vì \(\frac{a}{b+c+d}< 1\Rightarrow\frac{a}{b+c+d}< \frac{a+c}{b+c+a+d}\)
\(\frac{b}{c+d+a}< 1\Rightarrow\frac{b}{b+c}< \frac{b+a}{a+b+c+d}\)
\(\frac{c}{b+c+d}< 1\Rightarrow\frac{c}{b+c+d}< \frac{c+b}{a+b+c+d}\)
\(\frac{d}{a+b+c}< 1\Rightarrow\frac{d}{a+b+c}< \frac{d+b}{a+b+c+d}\)
\(\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< \frac{a+c}{a+b+c+d}+\frac{b+a}{a+b+c+d}+\frac{c+d}{a+b+c+d}+\frac{d+b}{a+b+c+d}\)
\(\Rightarrow\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< \frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow1< \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< 2\)
Vậy a,b,c,d>0 thì \(1< \frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{b+c+d}+\frac{d}{a+b+c}< 2\left(đpcm\right)\)
Ta có :
\(\frac{a}{a+b+c}>\frac{a}{a+b+c+d}\)
\(\frac{b}{a+b+d}>\frac{b}{a+b+c+d}\)
\(\frac{c}{b+c+d}>\frac{c}{a+b+c+d}\)
\(\frac{d}{c+d+a}>\frac{d}{a+b+c+d}\)
\(\Rightarrow\)\(M=\frac{a}{a+b+c}+\frac{b}{a+b+d}+\frac{c}{b+c+d}+\frac{d}{c+d+a}>\frac{a+b+c+d}{a+b+c+d}=1\) ( cộng theo vế 4 đẳng thức trên )
\(\Rightarrow\)\(M>1\) \(\left(1\right)\)
Lại có : ( phần này áp dụng công thức \(\frac{a}{b}< \frac{a+m}{b+m}\) \(\left(\frac{a}{b}< 1;a,b,m\inℕ^∗\right)\) )
\(\frac{a}{a+b+c}< \frac{a+d}{a+b+c+d}\)
\(\frac{b}{a+b+d}< \frac{b+c}{a+b+c+d}\)
\(\frac{c}{b+c+d}< \frac{c+a}{a+b+c+d}\)
\(\frac{d}{c+d+a}< \frac{d+b}{a+b+c+d}\)
\(\Rightarrow\)\(M=\frac{a}{a+b+c}+\frac{b}{a+b+d}+\frac{c}{b+c+d}+\frac{d}{c+d+a}< \frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\) ( cộng theo vế 4 đẳng thức trên )
\(\Rightarrow\)\(M< 2\) \(\left(2\right)\)
Từ (1) và (2) suy ra đpcm : \(1< M< 2\)
Vậy \(1< M< 2\)
Chúc bạn học tốt ~