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Áp dụng BĐT Svác - xơ.
\(F=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)
\(=\frac{a^2}{ba+ca}+\frac{b^2}{cb+db}+\frac{c^2}{dc+ac}+\frac{d^2}{ad+bd}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{ba+ca+bd+db+dc+ac+ad+bd}\)(1)
Xét: \(\left(a+b+c+d\right)^2-2\left(ba+ca+bd+db+dc+ac+ad+bd\right)\)
\(=a^2+b^2+c^2+d^2-2bd-2ac\)
\(=\left(a-c\right)^2+\left(b-d\right)^2\ge0\)
=> \(\left(a+b+c+d\right)^2\ge2\left(ba+ca+bd+db+dc+ac+ad+bd\right)\)
=> \(\frac{\left(a+b+c+d\right)^2}{ba+ca+bd+db+dc+ac+ad+bd}\ge2\)(2)
Từ ( 1); (2) => \(F\ge2\)
Dấu "=" xảy ra <=> a = b = c = d.
Ta có :
\(\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}\ge\frac{a-d}{a+b}\) (1)
\(\Leftrightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge0\)
\(\Leftrightarrow\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\ge4\)( Cộng mỗi phân số vs 1 )
\(\Leftrightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge4\) (2)
Với a ,b ,c ,d là các số dương , áp dụng BĐT Svacsơ , ta có :
\(\hept{\begin{cases}\frac{1}{b+c}+\frac{1}{d+a}\ge\frac{4}{a+b+c+d}\\\frac{1}{c+d}+\frac{1}{a+b}\ge\frac{4}{a+b+c+d}\end{cases}}\)
Suy ra : \(\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge\frac{4\left(a+c\right)+4\left(b+d\right)}{a+b+c+d}\)
\(\Leftrightarrow\left(2\right)\)\(\Leftrightarrow\left(1\right)\)( Điều cần CM )
\(Để\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge0\)
Thì \(\frac{a-b}{b+c}+1+\frac{b-c}{c+d}+1+\frac{c-d}{d+a}+1+\frac{d-a}{a+b}+1\ge4\)
\(\Leftrightarrow\frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\ge4\)
\(\Leftrightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\ge4\)(Cần phải chứng minh)
Ta có : \(\Leftrightarrow\left(a+c\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)+\left(b+d\right)\left(\frac{1}{c+d}+\frac{1}{a+b}\right)\)
\(\ge\left(a+c\right)\left(\frac{4}{a+b+c+d}\right)+\left(b+d\right)\left(\frac{4}{a+b+c+d}\right)=4\)(Áp dụng Cô-si dạng phân thức)
\(\Rightarrow\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge0\)(Đpcm)
Học tốt ~~
áp dụng bất đẳng thức Cauchy-schwaz
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}\)=\(\frac{16}{a+b+c+d}\)(đpcm)
Lời giải:
Điều kiện đề bài đã cho tương đương với:
\(\frac{a}{a+b}+\frac{b}{b+c}-1+\frac{c}{c+d}+\frac{d}{a+d}-1=0\)
\(\Leftrightarrow \frac{a}{a+b}-\frac{c}{b+c}+\frac{c}{c+d}-\frac{a}{a+d}=0\)
\(\Leftrightarrow a(\frac{1}{a+b}-\frac{1}{a+d})+c(\frac{1}{d+c}-\frac{1}{b+c})=0\)
\(\Leftrightarrow \frac{a(d-b)}{(a+b)(a+d)}+\frac{c(b-d)}{(d+c)(b+c)}=0\)
\(\Leftrightarrow (d-b)(\frac{a}{(a+b)(a+d)}-\frac{c}{(c+d)(c+b)})=0\)
\(\Leftrightarrow \frac{(d-b)(a-c)(bd-ac)}{(a+b)(a+d)(c+d)(c+b)}=0\)
\(\Rightarrow (d-b)(a-c)(bd-ac)=0\)
Mà $a,b,c,d$ đôi một khác nhau nên suy ra $bd-ac=0$
$\Rightarrow bd=ac$
$\Rightarrow abcd=(bd)^2$ là số chính phương với mọi $a,b,c,d$ nguyên dương.
Ta có đpcm.
Lời giải:
Điều kiện đề bài đã cho tương đương với:
\(\frac{a}{a+b}+\frac{b}{b+c}-1+\frac{c}{c+d}+\frac{d}{a+d}-1=0\)
\(\Leftrightarrow \frac{a}{a+b}-\frac{c}{b+c}+\frac{c}{c+d}-\frac{a}{a+d}=0\)
\(\Leftrightarrow a(\frac{1}{a+b}-\frac{1}{a+d})+c(\frac{1}{d+c}-\frac{1}{b+c})=0\)
\(\Leftrightarrow \frac{a(d-b)}{(a+b)(a+d)}+\frac{c(b-d)}{(d+c)(b+c)}=0\)
\(\Leftrightarrow (d-b)(\frac{a}{(a+b)(a+d)}-\frac{c}{(c+d)(c+b)})=0\)
\(\Leftrightarrow \frac{(d-b)(a-c)(bd-ac)}{(a+b)(a+d)(c+d)(c+b)}=0\)
\(\Rightarrow (d-b)(a-c)(bd-ac)=0\)
Mà $a,b,c,d$ đôi một khác nhau nên suy ra $bd-ac=0$
$\Rightarrow bd=ac$
$\Rightarrow abcd=(bd)^2$ là số chính phương với mọi $a,b,c,d$ nguyên dương.
Ta có: \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\)
\(>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}\)
\(=\frac{a+b+c+d}{a+b+c+d}=1\)
Tương tự ta cũng chứng minh được \(\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}>1\)
mà \(\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\right)+\left(\frac{b}{a+b}+\frac{c}{b+c}+\frac{d}{c+d}+\frac{a}{d+a}\right)\)
\(=\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+d}{c+d}+\frac{d+a}{d+a}=4\)
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}\)là số nguyên
do đó \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\)
\(\Leftrightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b}{a+b}-\frac{b}{b+c}+\frac{d}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow b\left(c+d\right)\left(d+a\right)-d\left(a+b\right)\left(b+c\right)=0\)(vì \(a\ne c\))
\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow ac=bd\)(vì \(b\ne d\))
Khi đó \(abcd=ac.ac=\left(ac\right)^2\)là số chính phương.
Ta có :
\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}=2\)
\(\Rightarrow1-\frac{a}{a+b}-\frac{b}{b+c}+1-\frac{c}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b}{a+b}-\frac{b}{b+c}+\frac{d}{c+d}-\frac{d}{d+a}=0\)
\(\Leftrightarrow\frac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\frac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow b\left(c+d\right)\left(d+a\right)+d\left(a+b\right)\left(b+c\right)=0\)( vì c khác a )
\(\Leftrightarrow abc-acd+bd^2-b^2d=0\)
\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow ac-bd=0\)
\(\Leftrightarrow ac=bd\)
\(\Rightarrow abcd=\left(ac\right)\left(bd\right)=\left(ac\right)^2\)
Vậy ......................................
A=\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}+\left(\frac{b}{b+c}+\frac{c}{c+d}+\frac{d}{d+a}+\frac{a}{a+b}\right)\)\(\ge4\)
B=\(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}+\left(\frac{c}{b+c}+\frac{d}{c+b}+\frac{a}{d+a}+\frac{b}{a+b}\right)\)\(\ge4\)
A+B=2M+2\(\ge\)8 (M là biểu thức cần chứng minh)
M\(\ge\)2 <=>a=b=c=d
Ta có
\(\frac{a}{b+c}\ge\frac{a+a+d}{a+b+c+d}\)
\(\frac{b}{c+d}\ge\frac{b+b+a}{a+b+c+d}\)
\(\frac{c}{d+a}\ge\frac{c+c+b}{a+b+c+d}\)
\(\frac{d}{a+b}\ge\frac{d+d+c}{a+b+c+d}\)
=> \(\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)> \(\frac{a+a+d+b+b+a+c+c+b+d+d+c}{a+b+c+d}\)=\(\frac{2a+2b+2c+2d}{a+b+c+d}\)= 2
Chúc bạn học tốt!