Giải:
\(\Leftrightarrow\left(\dfrac{m^2}{4}-mn+n^2\right)+\left(\dfrac{m^2}{4}-mp+p^2\right)+\left(\dfrac{m^2}{4}-mq+q^2\right)+\left(\dfrac{m^2}{4}-m+1\right)\ge0\)
\(\Leftrightarrow\left(\dfrac{m}{2}-n\right)^2+\left(\dfrac{m}{2}-p\right)^2+\left(\dfrac{m}{2}-q\right)^2+\left(\dfrac{m}{2}-1\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{m}{2}-n=0\\\dfrac{m}{2}-p=0\\\dfrac{m}{2}-q=0\\\dfrac{m}{2}-1=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}n=\dfrac{m}{2}\\p=\dfrac{m}{2}\\q=\dfrac{m}{2}\\m=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m=2\\n=p=q=1\end{matrix}\right.\)
m2+n2+p2+q2+1\(\ge\)m(n+p+q+1)(*)
nhân cả hai vế cho 4 ta được
(*)<=>(m2-4mn+4n2)+(m2-4mp+4p2)+(m2-4mq+4q2)+(m2-4m+4)\(\ge0\)
<=>(m-2n)2+(m-2p)2+(m-2q)2+(m-1)2\(\ge0\)
luôn đúng=>điều phải chứng minh
