b, \(\frac{a+b}{a+b+c}>\frac{a+b}{a+b+c+d}\); \(\frac{b+c}{b+c+a}>\frac{b+c}{a+b+c+d}\)

 \(\frac{c+d}{c+d+a}>\frac{c+d}{a+b+c+d};\frac{d+a}{a+d+b}>\frac{a+d}{a+b+c+d}\)

Cộng các bĐT trên

=> \(B>\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)

Ta  có Với \(0< \frac{x}{y}< 1\)

=> \(\frac{x}{y}< \frac{x+z}{y+z}\)

Áp dụng ta có 

\(B>\frac{a+b+d}{a+b+c+d}+...+\frac{d+a+c}{a+b+c+d}=3\)

Vậy 2<B<3

Áp dụng tính chất.......

a/b=b/c=c/d=a+b+c/b+c+d suy ra (a/b)^3=(b/c)^3=(c/d)^3=(a+b+c)^3/(b+c+d)^3(1)

a/b= b/c=c/dsuy ra a^3/b^3=b^3/c^3=c^3/d^3(2)

Áp dụng tính chất .....

a^3/b^3=b^3/c^3=c^3/d^3=a^3+b^3+c^3/b^3+c^3+d^3 (3)

Từ 1,2 và 3 suy ra :a^3+b^3+c^3/b^3+c^3+d^3=(a+b+c)^3/(b+c+d)^3

25361

\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b-c}{b+c-d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b-c}{b+c-d}\right)^3\)

Mà \(\left(\frac{a}{b}\right)^3=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\)

=>\(\left(\frac{a+b-c}{b+c-d}\right)^3=\frac{a}{d}\Rightarrow\frac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}=\frac{a}{d}\Rightarrow\frac{\left(a+b-c\right)^3}{a}=\frac{\left(b+c-d\right)^3}{d}\) (đpcm)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)

\(\Rightarrow k=\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\Rightarrow k^3=\left(\frac{a+b+c}{b+c+d}\right)^3\left(1\right)\)

Mà \(k^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\left(2\right)\)

Từ (1),(2)\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\left(đpcm\right)\)

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

\(\Rightarrow\)\(\frac{a}{b}\)³=\(\frac{b}{c}\)³=\(\frac{c}{d}\)³=\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)=\(\left(\frac{a+b+c}{b+c+d^{ }}\right)\)³

\(\Rightarrow\)\(\left(\frac{a+b+c}{b+c+d^{ }}\right)\)³=\(\frac{a}{b}\).\(\frac{b}{c}\).\(\frac{c}{d}\)=\(\frac{a}{d}\)

\(\Rightarrow\)đpcm

Có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)(Tính chất dãy tỉ số bằng nhau)

=> \(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{c^2}{d^2}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}=\frac{abc}{bcd}=\frac{a}{d}\)

=> \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)

=> Đpcm

Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\frac{abc}{bcd}=\frac{a}{d}\)( theo TC dãy TSBN)

Mà:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)(theo TC dãy TSBN)

=>\(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\left(=\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3\right)\)(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