con chó\

Ta có : 

\(P=a+2b=\left(a-2\right)+2\left(b+3\right)-4\)

\(\Rightarrow P+4=\left(a-2\right)+2\left(b+3\right)\)

\(\Rightarrow\left(P+4\right)^2=\left(\left(a-2\right)+2\left(b+3\right)\right)^2\le\left(1^2+2^2\right)\left(\left(a-2\right)^2+\left(b+3\right)^2\right)\)

\(=25\)

\(\Rightarrow-5\le P+4\le5\)

\(\Rightarrow P\ge-9\)

Dấu " = " xảy ra khi \(\frac{a-2}{1}=\frac{b+3}{2},\left(a-2\right)^2+\left(b+3\right)^2=5\)

\(\Rightarrow a-2=-1,b+3=-2\Rightarrow a=1,b=-5\)

Chứng minh \(\frac{1}{\left(x+y\right)\left(x+z\right)}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)

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

                                                                \(\le\frac{1}{4}.\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)\)

                                                                  \(\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

 tương tự cộng zế zới zế ta đc

\(\frac{1}{2xx+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)

                                                                                      \(\le\frac{1}{16}.4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)(dpcm

Ta có: \(6x^2+8xy+11y^2=2\left(x-y\right)^2+\left(2x+3y\right)^2\ge\left(2x+3y\right)^2\)

Tương tự: \(6y^2+8yz+11z^2\ge\left(2y+3z\right)^2\)

\(6z^2+8zx+11x^2\ge\left(2z+3x\right)^2\)

=> \(P\le\frac{x^2+3xy+y^2}{2x+3y}+\frac{y^2+3yz+z^2}{2y+3z}+\frac{z^2+3zx+x^2}{2z+3x}\)

=> \(4P\le\frac{4x^2+12xy+4y^2}{2x+3y}+\frac{4y^2+12yz+4z^2}{2y+3z}+\frac{4z^2+12zx+4x^2}{2z+3x}\)

\(=\frac{\left(2x+3y\right)^2-5y^2}{2x+3y}+\frac{\left(2y+3z\right)^2-5z^2}{2y+3z}+\frac{\left(2z+3x\right)^2-5x^2}{2z+3x}\)

\(=5\left(x+y+z\right)-5\left(\frac{y^2}{2x+3y}+\frac{z^2}{2y+3z}+\frac{x^2}{2z+3x}\right)\)

\(\le5\left(x+y+z\right)-5.\frac{\left(x+y+z\right)^2}{5\left(x+y+z\right)}=4\left(x+y+z\right)\)

Lại có: \(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\)với mọi x; y; z

=> \(4P\le4.\sqrt{9}=12\)

=> \(P\le3\)

Dấu "=" xảy ra <=> x = y = z = 1

Vậy max P = 3 đạt tại x = y = z = 1.