cho a,b,c,d>0 va abcd=1. chứng minh rằng: \(a^2+b^2+c^2+d^2+a\left(b+c\right)+b\left(c+d\right)+d\left(c+a\right)\ge10\)
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\(\left\{{}\begin{matrix}ab+ac+bc+bd+cd+da\ge4\sqrt[6]{ab.ac.bc.bd.cd.da}=6.\sqrt{abcd}=6\\a^2+b^2+c^2+d^2\ge4\sqrt[4]{a^2b^2c^2d^2}=4.\sqrt{abcd}=4\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\left(2\right)\end{matrix}\)
(1) cộng (2) => dpcm
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Ta có: \(a^2+b^2+c^2+d^2\ge4\sqrt[4]{\left(abcd\right)^2}=4\)(AM-GM) (abcd=1)
Lại có: \(a\left(b+c\right)+b\left(c+d\right)+c\left(d+a\right)+d\left(a+b\right)\)
\(=ab+ac+bc+bd+cd+ac+ad+bd\)
\(\ge8\sqrt[8]{\left(abcd\right)^4}=8\)(AM-GM)
Từ đó:
\(a^2+b^2+c^2+d^2+a\left(b+c\right)+b\left(c+d\right)+c\left(d+a\right)+d\left(a+b\right)\ge4+8=12\)
=> ĐPCM. Dấu "=" xảy ra <=> a=b=c=d=1.
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Theo giả thiết \(a^2+b^2+c^2+d^2=1\Rightarrow0< a,b,c,d< 1\)
Ta có: \(2\left(1-a\right)\left(1-b\right)=2-2\left(a+b\right)+2ab=a^2+b^2+c^2+d^2+1\)\(-2a-2b+2ab-2cd+2cd=\left(a+b-1\right)^2+\left(c-d\right)^2+2cd\ge2cd\)
\(\Rightarrow\left(1-a\right)\left(1-b\right)\ge cd\)(*)
Tương tự ta có: \(\left(1-c\right)\left(1-d\right)\ge ab\)(**)
Nhân theo từng vế cùng chiều của hai BĐT (*) và (**), ta được: \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\left(1-d\right)\ge abcd\)
Đẳng thức xảy ra khi \(a=b=c=d=\frac{1}{2}\)
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Áp dụng BĐT Bunhiacopxki:
\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\ge\sqrt{\left(ac+bc\right)^2}=ac+bc\)
CMTT : \(\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ad+bd\)
Ta có :\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ac+bc+ad+bd=\left(a+b\right)\left(c+d\right)\)
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Theo đề bài ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\) ( 1 )
Theo tính chất dãy tỉ số bằng nhau ta có :
\(k=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(k^2=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( 2 )
Mà từ ( 1 ) = > \(k^2=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\) ( 3 )
Từ ( 2 ) , ( 3 )
= > \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( đpcm )
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)
Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)
Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)
\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)
Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)
Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*
\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)
\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)
Đẳng thức xảy ra khi a = b = c
P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:
1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)
\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)
bài 2 xem có ghi nhầm ko
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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk\)
Ta có: \(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\left(1\right)\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{bk-b}{dk-d}=\dfrac{b\left(k-1\right)}{d\left(k-1\right)}=\dfrac{b}{d}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\\ \dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
nhan ra het roi dung cosi
1 cai la ra lien
\(abcd=1;ab=\frac{1}{cd};ad=\frac{1}{bc};ac=\frac{1}{bd}\)
Ta có : \(a^2+b^2+c^2+d^2+a\left(b+c\right)+b\left(c+d\right)+d\left(c+a\right)\)
\(=a^2+b^2+c^2+d^2+ab+ac+bc+bd+dc+ad\)
\(=a^2+b^2+c^2+d^2+\frac{1}{cd}+cd+\frac{1}{bd}+bd+\frac{1}{bc}+bc\)
\(\ge4\sqrt[4]{abcd}+2\sqrt{\frac{1}{cd}.cd}+2\sqrt{\frac{1}{bd}.bd}+2\sqrt{\frac{1}{bc}.bc}\)(Cauchy)
\(=4+2+2+2=10\)(đpcm)
Dấu"=" xảy ra \(\Leftrightarrow a=b=c=1\)