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 \(\Leftrightarrow a=b=c=1\)

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.

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 \(a=b=c=d=\frac{1}{2}\)

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 :\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}+\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\ge ac+bc+ad+bd=\left(a+b\right)\left(c+d\right)\)

Áp dụng BĐT Bunhiacopxki:

(�2+�2)(�2+�2)≥(��+��)2=��+��(a2+c2)(b2+c2)​≥(ac+bc)2​=ac+bc

CMTT : (�2+�2)(�2+�2)≥��+��(a2+d2)(b2+d2)​≥ad+bd

Ta có :(�2+�2)(�2+�2)+(�2+�2)(�2+�2)≥��+��+��+��=(�+�)(�+�)(a2+c2)(b2+c2)​+(a2+d2)(b2+d2)​≥ac+bc+ad+bd=(a+b)(c+d)

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 \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\Rightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\\ \dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

Cách giải:

1+1=3

6-6=0

9-9=0

Vậy => 6-6=9-9

(3-3)+(3-3) = 3x3 - 3x3

(1+1)=3

1+1=3

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 )

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

3a biến đổi tí là xong

b tuong tự bài 1