+) m= -1F(x;y) = $(x- y+1)^{2} +(-x +y+1)^{2}$ =$(x-y +1)^{2} +(x-y-1 )^{2}$ đặt t= x-y+1 khi đó F(x;y) = $2t^{2} - 4t +4 =2(t -1)^{2} +2\geq2$ dấu "=' $\Leftrightarrow$ t=1 $\Leftrightarrow$ y=x - 2 +) m $\neq$ -1 $ F\geq 0 dấu "=" \Leftrightarrow \begin{cases}x- y+1= 0\\ mx +y+m+2=0 \end{cases}$ $\Leftrightarrow \begin{cases}y=x+1 \\ mx +y +m+2=0 \end{cases}$ $\Leftrightarrow \begin{cases}x=\frac{-m -3}{m+1} \\ y=\frac{-2}{m +1} \end{cases}$

+) m= -1F(x;y) = $(x- y+1)^{2} +(-x +y+1)^{2}$ =$(x-y +1)^{2} +(x-y-1 )^{2}$ đặt t= x-y+1 khi đó F(x;y) = $2t^{2} - 4t +4 =2(t -1)^{2} +2\geq2$ dấu "=' $\Leftrightarrow$ t=1 $\Leftrightarrow$ y=x +) m $\neq$ -1 $ F\geq 0 dấu "=" \Leftrightarrow \begin{cases}x- y+1= 0\\ mx +y+m+2=0 \end{cases}$ $\Leftrightarrow \begin{cases}y=x+1 \\ mx +y +m+2=0 \end{cases}$ $\Leftrightarrow \begin{cases}x=\frac{-m -3}{m+1} \\ y=\frac{-2}{m +1} \end{cases}$

+) m= -1F(x;y) = $(x- y+1)^{2} +(-x +y+1)^{2}$ =$(x-y +1)^{2} +(x-y-1 )^{2}$ đặt t= x-y+1 khi đó F(x;y) = $2t^{2} - 4t +4 =2(t -1)^{2} +2\geq2$ dấu "=' $\Leftrightarrow$ t=1 $\Leftrightarrow$ y=x - 2 +) m $\neq$ -1 $ F\geq 0 dấu "=" \Leftrightarrow \begin{cases}x- y+1= 0\\ mx +y+m+2=0 \end{cases}$ $\Leftrightarrow \begin{cases}y=x+1 \\ mx +y +m+1=0 \end{cases}$ $\Leftrightarrow \begin{cases}x=\frac{-m -2}{m+1} \\ y=\frac{-1}{m +1} \end{cases}$

+) m= -1F(x;y) = $(x- y+1)^{2} +(-x +y+1)^{2}$ =$(x-y +1)^{2} +(x-y-1 )^{2}$ đặt t= x-y+1 khi đó F(x;y) = $2t^{2} - 4t +4 =2(t -1)^{2} +2\geq2$ dấu "=' $\Leftrightarrow$ t=1 $\Leftrightarrow$ y=x - 2 +) m $\neq$ -1 $ F\geq 0 dấu "=" \Leftrightarrow \begin{cases}x- y+1= 0\\ mx +y+m+2=0 \end{cases}$ $\Leftrightarrow \begin{cases}y=x+1 \\ mx +y +m+2=0 \end{cases}$ $\Leftrightarrow \begin{cases}x=\frac{-m -3}{m+1} \\ y=\frac{-2}{m +1} \end{cases}$

ad $\frac{1}{x} +\frac{1}{y}$ $\geq \frac{4}{x+y}$ ta có $\frac{1}{4a+ b+c} \leq \frac{1}{4} (\frac{1}{2a +b} +\frac{1}{2a +c})$ có $\frac{1}{2a +b} \leq \frac{1}{9}(\frac{1}{a} +\frac{1}{a} +\frac{1}{b})$ TT $\Rightarrow \frac{1}{4a +b +c} \leq \frac{1}{36} (\frac{4}{a} +\frac{1}{b} +\frac{1}{c}$ Từ đó $\Rightarrow VT \leq \frac{1}{6} (\frac{1}{a} + \frac{1}{b} +\frac{1}{c})$ ta cần cm $\frac{1}{a}+ \frac{1}{b} +\frac{1}{c} \leq 1$ ta thấy 3($\frac{1}{a^{2}} + \frac{1}{b^{2}}+ \frac{1}{c^{2}}$) $\geq$ (\frac{1}{a}+ \frac{1}{b} +\frac{1}{c})^2 $\Leftrightarrow 4( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c})^{2} \leq \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$ +3 (do gt) $\Leftrightarrow \frac{1}{a} +\frac{1}{b} +\frac{1}{c} \leq$ 1 $\Rightarrow$ đpcm dấu "=" $\Leftrightarrow$ a=b=c=3

ad $\frac{1}{x} +\frac{1}{y}$ $\geq \frac{4}{x+y}$ ta có $\frac{1}{4a+ b+c} \leq \frac{1}{4} (\frac{1}{2a +b} +\frac{1}{2a +c})$ có $\frac{1}{2a +b} \leq \frac{1}{9}(\frac{1}{a} +\frac{1}{a} +\frac{1}{b})$ TT $\Rightarrow \frac{1}{4a +b +c} \leq \frac{1}{36} (\frac{4}{a} +\frac{1}{b} +\frac{1}{c})$ Từ đó $\Rightarrow VT \leq \frac{1}{6} (\frac{1}{a} + \frac{1}{b} +\frac{1}{c})$ ta cần cm $\frac{1}{a}+ \frac{1}{b} +\frac{1}{c} \leq 1$ ta thấy 3($\frac{1}{a^{2}} + \frac{1}{b^{2}}+ \frac{1}{c^{2}}$) $\geq (\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c})^{2}$ $\Leftrightarrow 4( \frac{1}{a}+ \frac{1}{b}+ \frac{1}{c})^{2} \leq \frac{1}{a} +\frac{1}{b} +\frac{1}{c}$ +3 (do gt) $\Leftrightarrow \frac{1}{a} +\frac{1}{b} +\frac{1}{c} \leq$ 1 $\Rightarrow$ đpcm dấu "=" $\Leftrightarrow$ a=b=c=3