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\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow b\left(c-a\right)\left(a+b\right)\left(b+c\right)-d\left(c-a\right)\left(c+d\right)\left(d+a\right)=0\)
\(\Leftrightarrow b\left(a+b\right)\left(b+c\right)-d\left(c+d\right)\left(d+a\right)=0\)
\(\Leftrightarrow bad+bd^2+bca+bcd-dab-dac-db^2-cbd=0\)
\(\Leftrightarrow bca-dca+bd^2-db^2=0\)
\(\Leftrightarrow\left(b-d\right)\left(ca-bd\right)=0\)
\(\Rightarrow ca=bd\Rightarrow abcd=bd^2\)
Ta có: \(A=\frac{a-d}{d+b}+\frac{d-b}{b+c}+\frac{b-c}{c+a}+\frac{c-a}{a+d}\)
\(\Leftrightarrow A+4=\frac{a-d}{d+b}+1+\frac{d-b}{b+c}+1+\frac{b-c}{c+a}+1+\frac{c-a}{a+d}+1\)
\(\Leftrightarrow A+4=\frac{a+b}{d+b}+\frac{d+c}{b+c}+\frac{b+a}{c+a}+\frac{c+d}{a+d}\)
\(\Leftrightarrow A+4=\left(a+b\right)\left(\frac{1}{d+b}+\frac{1}{a+c}\right)+\left(c+d\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\)
Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{xy}\)với mọi x,y>0
Ta có: \(A+4\ge\frac{4\left(a+b\right)}{a+b+c+d}+\frac{4\left(d+c\right)}{a+b+c+d}\)
\(A+4\ge\frac{4\left(a+b+c+d\right)}{a+b+c+d}=4\)
\(A\ge0\)(dpcm)
Ta có : \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
1) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{b}{d}=\frac{a}{c}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
đpcm
2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=\frac{a+b}{c+d}\)
\(\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)
đpcm
\(bdt\Leftrightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}+\frac{d-a}{a+b}\ge0\)
\(\Leftrightarrow\left(\frac{a-b}{b+c}+1\right)+\left(\frac{b-c}{c+d}+1\right)+\left(\frac{c-d}{d+a}+1\right)+\left(\frac{d-a}{a+b}+1\right)\ge4\)
\(\Leftrightarrow\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{a+c}{d+a}+\frac{b+d}{a+b}\ge4\)
\(\Leftrightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge4\)(*)
Theo Cauchy-Schwarz:
\(\frac{1}{b+c}+\frac{1}{d+a}\ge\frac{4}{a+b+c+d};\frac{1}{c+d}+\frac{1}{a+b}\ge\frac{4}{a+b+c+d}\)
Khi đó:\(\left(\cdot\right)\ge\left(a+c\right).\frac{4}{a+b+c+d}+\left(b+d\right).\frac{4}{a+b+c+d}=4\)
Đặt \(A=\frac{\left(a+b+c+d\right)\left(a+b+c\right)\left(a+b\right)}{abcde}\)
\(\Rightarrow16A=\frac{\left(a+b+c+d+e\right)^2\left(a+b+c+d\right)\left(a+b+c\right)\left(a+b\right)}{abcde}\)
Áp dụng AM-GM ta có:
\(\Rightarrow16A\ge\frac{4e\left(a+b+c+d\right)^2\left(a+b+c\right)\left(a+b\right)}{abcde}\)
\(\Rightarrow16A\ge\frac{4e.4d\left(a+b+c\right)^2\left(a+b\right)}{abcde}\)
\(\Rightarrow16A\ge\frac{4e.4d.4c\left(a+b\right)^2}{abcde}\)
\(\Rightarrow16A\ge\frac{4e.4d.4c.4ab}{abcde}\)
\(\Rightarrow A\ge16\)
Dấu "=" xảy ra khi đồng thời:
\(\text{a+b+c+d+e=4, a+b+c+d=e, a+b+c=d, a+b=c, a=b}\)
\(\Rightarrow e=2,d=1,c=\frac{1}{2},a=\frac{1}{4},b=\frac{1}{4}\)
a/ Biến đổi tương đương:
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
Vậy BĐT được chứng minh
b/ \(VT=\frac{a-d}{b+d}+1+\frac{d-b}{b+c}+1+\frac{b-c}{a+c}+1+\frac{c-a}{a+d}+1-4\)
\(VT=\frac{a+b}{b+d}+\frac{c+d}{b+c}+\frac{a+b}{a+c}+\frac{c+d}{a+d}-4\)
\(VT=\left(a+b\right)\left(\frac{1}{b+d}+\frac{1}{a+c}\right)+\left(c+d\right)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)-4\)
\(\Rightarrow VT\ge\left(a+b\right).\frac{4}{b+d+a+c}+\left(c+d\right).\frac{4}{b+c+a+d}-4\)
\(\Rightarrow VT\ge\frac{4}{\left(a+b+c+d\right)}\left(a+b+c+d\right)-4=4-4=0\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=d\)
TH1 : \(a-b=c-d=0\)
\(\Rightarrow a=b;c=d\)
\(\Rightarrow a+c=b+d\)
TH2 :\(a-b\ne0;c-d\ne0\)
\(\frac{a-b}{b-c}=\frac{c-d}{d-a}\)
\(\Rightarrow\left(a-b\right)\left(d-a\right)=\left(b-c\right)\left(c-d\right)\)
\(\Rightarrow ad-a^2-bd+ab=bc-bd-c^2+cd\)
\(\Rightarrow ad-a^2+ab=bc-c^2+cd\)
\(\Rightarrow a\left(d-a+b\right)=c\left(b-c+d\right)\)
Với \(d-a+b=b-c+d=0\)
\(\Rightarrow d-a+b-\left(d+b\right)=\left(b-c+d\right)-\left(d+b\right)\)
\(\Rightarrow a=c\)
Với \(d-a+b\ne0;b-c+d\ne0\)
\(\Rightarrow a=c\)
Vậy ...