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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: \(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=k+1\)
\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=k+1\)
=>\(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a-c}{c}=\dfrac{bk-dk}{dk}=\dfrac{b-d}{d}\)
b: \(\dfrac{a+b}{c+d}=\dfrac{bk+b}{dk+d}=\dfrac{b}{d}\)
\(\dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b}{d}\)
Do đó: \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{1}{a}+\frac{1}{b}\right)(a+b)\ge (1+1)^2\)
\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}\geq \frac{4}{a+b}\)
\(\Rightarrow \frac{c}{a}+\frac{c}{b}\geq \frac{4c}{a+b}\)
Hoàn toàn tương tự: \(\frac{a}{b}+\frac{a}{c}\geq \frac{4a}{b+c}; \frac{b}{a}+\frac{b}{c}\geq \frac{4b}{a+c}\)
Cộng theo vế các BĐT thu được:
\(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\geq 4\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\) (đpcm)
Dấu bằng xảy ra khi $a=b=c$
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)
\(\Rightarrow\left\{{}\begin{matrix}ad-bc< 0\\bd>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}ad< bc\\bd>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{bd}{ad}>\dfrac{bd}{bc}\)
\(\Rightarrow\dfrac{b}{a}>\dfrac{d}{c}\)
\(\rightarrowđpcm\)
Từ \(p=\dfrac{a+b+c}{2}\Rightarrow2p=a+b+c\)
Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\) ta có:
\(\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{p-a+p-b}=\dfrac{4}{2q-a-b}\)
\(=\dfrac{4}{a+b+c-a-b}=\dfrac{4}{c}\). Tương tự cho 2 BĐT còn lại:
\(\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{a};\dfrac{1}{p-c}+\dfrac{1}{p-a}\ge\dfrac{4}{b}\)
Cộng theo vế 3 BĐT trên ta có:
\(2\left(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\right)\ge4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
Đẳng thức xảy ra khi \(a=b=c\)
Ta có:
a/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a}{3b}=\dfrac{2c}{2d}=\dfrac{3a+2c}{3b+2d}\)
b/ \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{-2a}{-2b}=\dfrac{7c}{7d}=\dfrac{-2a+7c}{-2b+7d}\)
PS: Xong
\(\dfrac{a}{b}< \dfrac{a+c}{b+c}\)
\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)
\(\Leftrightarrow ab+ac< ba+bc\)
\(\Leftrightarrow ac< bc\)
\(\Leftrightarrow a< b\)(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
\(\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}\)\(\Leftrightarrow\dfrac{a+b+c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{2\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)