Ta dễ dàng chứng minh BĐT

\(x^4+y^4\ge x^3y+xy^3\)

\(\Rightarrow2\left(x^4+y^4\right)\ge x^4+y^4+x^3y+xy^3=\left(x+y\right)\left(x^3+y^3\right)\)

\(\Rightarrow\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)

Chứng minh tương tự, cộng theo vế, ta có:

\(\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^3+x^3}\ge\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=\frac{2\left(x+y+z\right)}{2}=2\)

Dấu "=" xảy ra khi x=y=z=1/3

Ta chứng minh \(x^4+y^4\ge x^3y+xy^3\)

\(\Leftrightarrow x^3\left(x-y\right)-y^3\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left[\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}\right]\ge0\)(luôn đúng)

Áp dụng vào bài toán ta có:

\(x^4+y^4\ge x^3y+xy^3\)\(\Rightarrow2\left(x^4+y^4\right)\ge x^4+y^4+x^3y+xy^3\)\(=\left(x^3+y^3\right)\left(x+y\right)\)

\(\Rightarrow\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\).Tương tự ta cũng có:

\(\frac{y^4+z^4}{y^3+z^3}\ge\frac{y+z}{2};\frac{z^4+x^4}{z^3+x^3}\ge\frac{z+x}{2}\)

Cộng theo vế ta có: \(VT\ge\frac{x+y}{2}+\frac{y+z}{2}+\frac{z+x}{2}=x+y+z=1\)

Dấu = khi \(x=y=z=\frac{2008}{3}\)

sửa đề: z+4>0

Đặt a = x + 1 > 0 ; b = y + 1 > 0 ; c = z + 4 > 0

a + b + c = 6

\(A=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)

Theo Bất Đẳng Thức ta có: \(\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{4}{c}\ge\frac{4}{a+b}+\frac{4}{c}\ge\frac{16}{a+b+c}=\frac{8}{3}\)

\(\Rightarrow A\le\frac{1}{3}\)Đẳng thức xảy ra khi và chỉ khi \(\hept{\begin{cases}a=b\\a+b=c\\a+b+c=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b=\frac{3}{2}\\c=3\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{2}\\z=-1\end{cases}}}\)

Vậy MaxA = 1/3 khi \(\hept{\begin{cases}x=y=\frac{1}{2}\\z=-1\end{cases}}\)

PaiN: Nhưng x,y,z là các số thực dương thì sao z âm đc?

\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)

Đặt: \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=k\)

\(\Rightarrow x=k\)

     \(y=2k\)

     \(z=3k\)

Thay x = k , y = 2k , z = 3k vào biểu thức cần cm ,ta đc:

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=\left(k+2k+3k\right)\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)

\(=6k.\left(\frac{1}{k}+\frac{2}{k}+\frac{3}{k}\right)\)

\(=6k.\frac{6}{k}\)

\(=\frac{36k}{k}=36\)

=.= hok tốt!!

Đặt \(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=k\)

Do đó  \(x=k;y=2k;z=3k\)

Thay \(x=k;y=2k;z=3k\)vào \(\left(x+y+z\right).\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)\)ta có 

\(\left(k+2k+3k\right).\left(\frac{1}{k}+\frac{4}{2k}+\frac{9}{3k}\right)\)

\(=6k.\left(\frac{6}{6k}+\frac{12}{6k}+\frac{18}{6k}\right)\)

\(=6k.\frac{6+12+18}{6k}\)

\(=\frac{6k.\left(6+12+18\right)}{6k}\)

\(=36\)

Do đó \(\left(x+y+z\right).\left(\frac{1}{x}+\frac{4}{y}+\frac{9}{z}\right)=36\)

Ta có:

\(\frac{x}{1}=\frac{y}{2}=\frac{z}{3}=\frac{x+y+z}{1+2+3}=\frac{x+y+z}{6}\)(Tính chất dãy tỉ số bằng nhau

=>  \(\hept{\begin{cases}\frac{x}{1}=\frac{x+y+z}{6}\\\frac{y}{2}=\frac{x+y+z}{6}\\\frac{z}{3}=\frac{x+y+z}{6}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{x+y+z}{6}\\y=\frac{x+y+z}{3}\\z=\frac{x+y+z}{2}\end{cases}}\)

Đặt biểu thức cần chứng minh là A và x + y + z = k

=> \(\hept{\begin{cases}x=\frac{k}{6}\\y=\frac{k}{3}\\z=\frac{k}{2}\end{cases}}\)

=> A = \(k\left(\frac{1}{\frac{k}{6}}+\frac{4}{\frac{k}{3}}+\frac{9}{\frac{k}{2}}\right)\)

    A = \(k.\left(\frac{6}{k}+\frac{12}{k}+\frac{18}{k}\right)=k.\frac{36}{k}=36\)(đpcm)