tự túc lm đi

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 BĐT cosi ta có 

\(\frac{a^6}{b^3}+\frac{b^6}{c^3}+1\ge3\sqrt[3]{\frac{a^6.b^3}{c^3}}=\frac{3a^2b}{c}\)

\(\frac{b^6}{c^3}+\frac{c^6}{a^3}+1\ge\frac{3b^2c}{a}\)

\(\frac{c^6}{a^3}+\frac{a^6}{b^3}+1\ge\frac{3c^2a}{b}\)

Cộng 3 bĐt trên

=> \(2.VT+3\ge3\left(\frac{a^2b}{c}+\frac{b^2c}{a}+\frac{c^2a}{b}\right)=9\)

=> \(VT\ge3\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

Ta có \(1+\frac{a}{x}=1+\frac{x+y+z}{x}=\frac{2x+y+z}{x}\)

Áp dụng BĐT cosi \(x+x+y+z\ge4\sqrt[4]{x^2yz}\)

=> \(1+\frac{a}{x}\ge\frac{4\sqrt[4]{x^2yz}}{x}\)

Tương tự\(1+\frac{a}{y}\ge\frac{4\sqrt[4]{y^2xz}}{y}\); \(1+\frac{a}{z}\ge\frac{4\sqrt[4]{z^2yx}}{z}\)

=> \(Q\ge\frac{64.\sqrt[4]{x^4y^4z^4}}{xyz}=64\)

MinQ=64 khi \(x=y=z=\frac{a}{3}\)

Ta có \(A=\frac{3x+4}{x^2+1}\)

=> \(Ax^2-3x+A-4=0\)

+ \(A=0\)

=> \(x=-\frac{4}{3}\)

+ \(A\ne0\)=> \(x\ne-\frac{4}{3}\)

=> \(\Delta=9-4A\left(A-4\right)\ge0\)

=> \(4A^2-16A-9\le0\)

=> \(-\frac{1}{2}\le A\le\frac{9}{2}\)

=> \(MinA=-\frac{1}{2}\)khi x=-3(TM \(x\ne\frac{-4}{3}\))

\(Max=\frac{9}{2}\)khi \(x=\frac{1}{3}\)(TM \(x\ne-\frac{4}{3}\))