`m^2+n^2+p^2+q^2+1>=m(n+p+q+1)`
Ta có :
`m^2+n^2+p^2+q^2+1>=m(n+p+q+1)`
`<=>((m^2)/4-mn+n^2)+((m^2)/4-mp+p^2)+((m^2)/4-mq+q^2)+((m^2)/4-m+1)>=0`
`<=>(m/2-n)^2+(m/2-p)^2+(m/2-q)^2+(m/2-1)^2>=0` ( luôn đúng )
Dấu "=" xảy ra khi :
$\begin{cases}\dfrac{m}{2}-n=0\\\dfrac{m}{2}-p=0\\\dfrac{m}{2}-q=0\\\dfrac{m}{2}-1=0\end{cases}$`=>` $\begin{cases}n=\dfrac{m}{2}\\p=\dfrac{m}{2}\\q=\dfrac{m}{2}\\m=2\end{cases}$ `=>` $\begin{cases}m=2\\n=1\\q=1\\p=1\end{cases}$
Vậy `m^2+n^2+p^2+q^2+1>=m(n+p+q+1)-> ( đpcm)`
