a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)

\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)

\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)

\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)

\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)

Bài 1:ta có BĐt \(a^3+b^3\ge ab\left(a+b\right)\)vì nó tương đương với \(\left(a+b\right)\left(a-b\right)^2\ge0\)(luôn đúng với a,b>0)

Áp dụng vào bài toán:

\(\dfrac{a^3+b^3}{2ab}+\dfrac{b^3+c^3}{2bc}+\dfrac{c^3+a^3}{2ac}\ge\dfrac{ab\left(a+b\right)}{2ab}+\dfrac{bc\left(b+c\right)}{2bc}+\dfrac{ca\left(c+a\right)}{2ac}=a+b+c\)dấu = xảy ra khi a=b=c

bài 2:

cần chứng minh \(\dfrac{a-b}{b+c}+\dfrac{b-c}{c+d}+\dfrac{c-d}{d+a}+\dfrac{d-a}{a+b}\ge0\)

hay \(\dfrac{a-b}{b+c}+1+\dfrac{b-c}{c+d}+1+\dfrac{c-d}{d+a}+1+\dfrac{d-a}{a+b}+1\ge4\)

\(\Leftrightarrow\dfrac{a+c}{b+c}+\dfrac{b+d}{c+d}+\dfrac{c+a}{d+a}+\dfrac{d+b}{a+b}\ge4\)

xét \(VT=\left(a+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+d}\right)+\left(b+d\right)\left(\dfrac{1}{c+d}+\dfrac{1}{a+b}\right)\)

Áp dụng BĐT cauchy dạng phân thức:

\(\dfrac{1}{b+c}+\dfrac{1}{a+d}\ge\dfrac{4}{a+b+c+d};\dfrac{1}{c+d}+\dfrac{1}{a+b}\ge\dfrac{4}{a+b+c+d}\)

do đó \(VT\ge\dfrac{4\left(a+c\right)}{a+b+c+d}+\dfrac{4\left(b+d\right)}{a+b+c+d}=4\)

dấu = xảy ra khi a=b=c=d

Ta có:

\(\dfrac{2a+b}{a+b}+\dfrac{2c+d}{c+d}+\dfrac{2b+c}{b+c}+\dfrac{2d+a}{d+a}=6\)

⇔ \(\left(\dfrac{2a+b}{a+b}-1\right)+\left(\dfrac{2c+d}{c+d}-1\right)+\left(\dfrac{2b+c}{b+c}-1\right)+\left(\dfrac{2d+a}{d+a}-1\right)=2\)

⇔ \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)

⇔ \(\left(1-\dfrac{a}{a+b}\right)-\dfrac{b}{b+c}+\left(1-\dfrac{c}{c+d}\right)-\dfrac{d}{d+a}=0\)

⇔ \(\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)

⇔ \(\dfrac{b\left(b+c\right)-b\left(a+b\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(d+a\right)-d\left(c+d\right)}{\left(c+d\right)\left(d+a\right)}=0\)

⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)

⇔ \(\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}-\dfrac{d\left(c-a\right)}{\left(c+d\right)\left(d+a\right)}=0\)

⇔ \(\left(c-a\right)\left(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right)=0\)

⇒ \(\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}=0\)         \(\left(a\ne c\right)\)

⇒ \(b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)

⇔ \(\left(bc+bd\right)\left(d+a\right)-\left(ad+bd\right)\left(b+c\right)=0\)

⇔ \(bcd+abc+bd^2+abd-abd-acd-b^2d-bcd=0\)

⇔ \(abc+bd^2-acd-b^2d=0\)

⇔ \(ac\left(b-d\right)-bd\left(b-d\right)=0\)

⇔ \(\left(b-d\right)\left(ac-bd\right)=0\)

⇒ \(ac-bd=0\)       \(\left(b\ne d\right)\)

⇔ \(ac=bd\)

Khi đó:

\(A=abcd=\left(ac\right)^2\)

⇒ \(ĐPCM\)

\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\\ \Rightarrow\dfrac{b+c+d}{a}=\dfrac{a+c+d}{b}=\dfrac{a+b+d}{c}=\dfrac{a+b+c}{d}=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\\ \Rightarrow\left\{{}\begin{matrix}b+d+c=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=4a\\a+b+c+d=4b\\a+b+c+d=4c\\a+b+c+d=4d\end{matrix}\right.\\ \Rightarrow4a=4b=4c=4d\Rightarrow a=b=c=d\\ \Rightarrow P=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)

\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)

ab+c+d=ba+c+d=ca+b+d=da+b+c=a+b+c+d3(a+b+c+d)=13⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩b+c+d=3aa+c+d=3ba+b+d=3ca+b+c=3d⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩a+b+c+d=2aa+b+c+d=2ba+b+c+d=2ca+b+c+d=2d⇒2a=2b=2c=2d⇒a=b=c=d⇒A=a+aa+a+a+aa+a+a+aa+a+a+aa+a=1+1+1+1=4

TH1: \(a+b+c+d\ne0\)

\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)

\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)

\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+2d}{d}\)

\(\Rightarrow a=b=c=d\)

\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)

\(\Rightarrow P=1+1+1+1\)

\(\Rightarrow P=4\)

TH2: \(a+b+c+d=0\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)

\(P=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{d+a}{b+c}\)

\(\Rightarrow P=\dfrac{-\left(c+d\right)}{c+d}+\dfrac{-\left(d+a\right)}{d+a}+\dfrac{-\left(a+b\right)}{a+b}+\dfrac{-\left(b+c\right)}{b+c}\)

\(\Rightarrow P=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)\)

\(\Rightarrow P=-4\)

\(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a}{a+b}=\dfrac{c}{c+d}\)

\(\Rightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)

\(TH1:a+b+c+d\ne0\)

\(\dfrac{2a+b+c+d}{a}=\dfrac{a+2b+c+d}{b}=\dfrac{a+b+2c+d}{c}=\dfrac{a+b+c+2d}{d}\)

\(\Rightarrow\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1=\dfrac{a+b+c+2d}{d}-1\)

\(\Rightarrow\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)

\(\Rightarrow a=b=c=d\)

\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)

\(=1+1+1+1\)

\(=4\)

\(TH2:a+b+c+d=0\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=-\left(c+d\right)\\b+c=-\left(d+a\right)\\c+d=-\left(a+b\right)\\d+a=-\left(b+c\right)\end{matrix}\right.\)

\(M=\dfrac{a+b}{c+d}+\dfrac{b+c}{d+a}+\dfrac{c+d}{a+b}+\dfrac{a+d}{b+c}\)

\(=-\dfrac{c+d}{c+d}-\dfrac{d+a}{d+a}-\dfrac{a+b}{a+b}-\dfrac{b+c}{b+c}\)

\(=-1-1-1-1\)

\(=-4\)

Lời giải:

Áp dụng TCDTSBN:

$\frac{a+b+c-d}{d}=\frac{b+c+d-a}{a}=\frac{c+d+a-b}{b}=\frac{d+a+b-c}{c}$

$=\frac{a+b+c-d+b+c+d-a+c+d+a-b+d+a+b-c}{d+a+b+c}$

$=\frac{2(a+b+c+d)}{a+b+c+d}=2$
$\Rightarrow a+b+c-d=2d; b+c+d-a=2a; c+d+a-b=2b; d+a+b-c=2c$

$\Rightarrow a+b+c=3d; b+c+d=3a; c+d+a=3b; d+a+b=3c$

Khi đó:

\(P=\frac{a+b+c}{a}.\frac{b+c+d}{b}.\frac{c+d+a}{c}.\frac{a+b+d}{d}\\ =\frac{3d}{a}.\frac{3a}{b}.\frac{3b}{c}.\frac{3c}{d}=81\)