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Ta có :
\(\dfrac{a}{b}< \dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{b}-\dfrac{c}{d}< 0\)
\(\Rightarrow\dfrac{ad-bc}{bd}< 0\)
Mà \(bd>0\) (do b,d dương)
\(\Rightarrow\left\{{}\begin{matrix}ad-bc< 0\\bd>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ad< bc\\bd>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{bd}{ad}>\dfrac{bd}{bc}\)
\(\Rightarrow\dfrac{b}{a}>\dfrac{d}{c}\)
\(\rightarrowđpcm\)
\(\dfrac{a+b}{a+b+c}\)>\(\dfrac{a+b}{a+b+c+d}\)
\(\dfrac{b+c}{b+c+d}\)>\(\dfrac{b+c}{b+c+d+a}\)
\(\dfrac{c+d}{c+d+a}\)>\(\dfrac{c+d}{c+d+a+b}\)
\(\dfrac{d+a}{d+a+b}\)>\(\dfrac{d+a}{d+a+b+c}\)
cộng từng vế của bất đẳng thức lại với nhau ta được
\(\dfrac{a+b}{a+b+c}\)+\(\dfrac{b+c}{b+c+d}\)+\(\dfrac{c+d}{c+d+a}\)+\(\dfrac{d+a}{d+a+b}\)>\(\dfrac{a+b}{a+b+c+d}\)+\(\dfrac{b+c}{b+c+d+a}\)+\(\dfrac{c+d}{c+d+a+b}\)+\(\dfrac{d+a}{d+a+b+c}\)=\(\dfrac{2.\left(a+b+c+d\right)}{a+b+c+d}\)=2
Áp dụng BĐT Svacxơ:
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{cd}+\dfrac{1}{da}\ge\dfrac{4}{ab+bc+cd+da}\)
Áp dụng BĐT Cô-si:
\(\dfrac{4}{ab+bc+cd+da}\ge\dfrac{4}{a^2+b^2+c^2+d^2}\)
Ta cần c/m: \(\dfrac{4}{a^2+b^2+c^2+d^2}\ge a^2+b^2+c^2+d^2\)
\(\Rightarrow\left(a^2+b^2+c^2+d^2\right)^2\ge4\)
Áp dụng BĐT Svacxơ: \(\left(\dfrac{a^2}{1}+\dfrac{b^2}{1}+\dfrac{c^2}{1}+\dfrac{d^2}{1}\right)^2\ge\dfrac{\left(a+b+c+d\right)^{2^2}}{16}\)
mà a+b+c+d=4 nên: \(\dfrac{\left(a+b+c+d\right)^4}{16}\ge\dfrac{64}{16}=4=VP\)
Vậy ta có đpcm.
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+d}+\dfrac{d^2}{a+d}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{a+b+b+c+c+d+d+a}\)
\(=\dfrac{\left(a+b+c+d\right)^2}{2\left(a+b+c+d\right)}=\dfrac{a+b+c+d}{2}=\dfrac{1}{2}=VP\)
Đẳng thức xảy ra khi \(a=b=c=d=\dfrac{1}{4}\)
a: ad=bc
=>a/b=c/d=k
=>a=bk; c=dk
b: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)
a/b=bk/b=k
=>(a+c)/(b+d)=a/b
c: ad=bc
nên a/c=b/d
d: \(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=k+1\)
\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=k+1\)
=>\(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow ab+ac< ba+bc\)
\(\Leftrightarrow ac< bc\)
\(\Leftrightarrow a< b\)(đúng)
a)Áp dụng
\(\Rightarrow\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=2\left(1\right)\)
Lại có:\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{b+c+a}+\dfrac{c}{c+a+b}=1\left(2\right)\)
Từ (1) và (2)=> đpcm
Vì \(\dfrac{a}{b}< 1\Rightarrow a< b\Rightarrow ac< bc\Rightarrow ac+ab< bc+ab\Rightarrow a\left(b+c\right)< b\left(a+c\right)\Rightarrow\dfrac{a\left(b+c\right)}{b\left(b+c\right)}< \dfrac{b\left(a+c\right)}{b\left(b+c\right)}\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+c}\)a) ta có
\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}\)\(\Leftrightarrow\dfrac{a+b+c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)