\(x^2+36y^2\ge12xy\) ; \(x^2+36z^2\ge12zx\) ; \(30y^2+30z^2\ge60yz\)

Cộng vế với vế:

\(2x^2+66y^2+66z^2\ge12xy+12yz+60zx\)

\(\Leftrightarrow2\left(x^2+33y^2+33z^2\right)\ge12\left(xy+yz+5zx\right)=204\)

\(\Rightarrow P\ge102\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x^2=36y^2=36z^2\\xy+zx+5yz=17\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=z^2=1\end{matrix}\right.\)

\(2=3\sqrt{xy}+2\sqrt{xz}\le\dfrac{3}{2}\left(x+y\right)+x+z\)

\(\Rightarrow5x+3y+2z\ge4\)

\(A=5\left(\dfrac{xy}{z}+\dfrac{xz}{y}\right)+3\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+2\left(\dfrac{xz}{y}+\dfrac{yz}{x}\right)\)

\(A\ge5.2x+3.2y+2.2z=2\left(5x+3y+2z\right)\ge8\)

\(A_{min}=8\) khi \(x=y=z=\dfrac{2}{5}\)

Thêm điều kiện x; y; z > 0

B1: Tìm điểm rơi 

B2: Dùng cô - si

\(S=3\left(x^2+y^2\right)+z^2=\left(2x^2+\frac{1}{2}z^2\right)+\left(2y^2+\frac{1}{2}z^2\right)+\left(x^2+y^2\right)\)

\(\ge2.\sqrt{x^2z^2}+2.\sqrt{y^2z^2}+2.\sqrt{x^2y^2}\)

\(=2\left(xy+yz+zx\right)=2\)

Dấu "=" xảy ra <=> \(x=y=\frac{1}{\sqrt{5}};z=\frac{2}{\sqrt{5}}\)

Ta có : \(x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}.2.\left(x^2+y^2+z^2-xy-yz-zx\right)\)

\(=\frac{1}{2}\left[\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\right]\ge0\)\(\Rightarrow x^2+y^2+z^2\ge xy+yz+xz\)

Đẳng thức xảy ra khi \(x=y=z\)

\(xy+yz+xz\le x^2+y^2+z^2\le3\)

\(\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+xz}\ge\frac{9}{3+xy+yz+zx}\ge\frac{9}{3+3}=\frac{3}{2}\)

Dấu"=" xảy ra khi x=y=z=1

Vậy...

MIn=3/2 khi x=y=z=1

\(P\ge\frac{9}{1+xy+1+yz+1+zx}=\frac{9}{3+\left(xy+yz+zx\right)}\)

Mà \(xy+yz+zx\le x^2+y^2+z^2\le3\)

\(P\ge\frac{9}{3+3}=\frac{3}{2}\)

Dấu bằng xảy ra khi x=y=z=1