Question

Chứng minh với mọi m , n , q ,p ta đều có :

m² + n² + p² + q² +1 \(\ge\) m(n +p +q +1 )

Answer 1

Ta có:

m²+n²+p²+q²+1-mn+mp+mq+m

\(=\left(\dfrac{m^2}{4}-mn+n^2\right)+\left(\dfrac{m^2}{4}-mp+p^2\right)+\left(\dfrac{m^2}{4}-mq+q^2\right)+\left(\dfrac{m^2}{4}-m+1\right)\)

\(=\left(\dfrac{m}{2}-n\right)^2+\left(\dfrac{m}{2}-p\right)^2+\left(\dfrac{m}{2}-q\right)^2+\left(\dfrac{m}{2}-1\right)^2\)

mà \(\left(\dfrac{m}{2}-n\right)^2\ge0;\left(\dfrac{m}{2}-p\right)^2\ge0;\left(\dfrac{m}{2}-q\right)^2\ge0;\left(\dfrac{m}{2}-1\right)^2\ge0\)

=> \(\left(\dfrac{m}{2}-n\right)^2+\left(\dfrac{m}{2}-p\right)^2+\left(\dfrac{m}{2}-q\right)^2+\left(\dfrac{m}{2}-1\right)^2\ge0\)

<=> m²+n²+p²+q²+1-mn+mp+mq+m \(\ge0\)

<=> m²+n²+p²+q²+1\(\ge\) mn+mp+mq+m

<=> m²+n²+p²+q²+1\(\ge\) m(n+p+q+1)

Vậy m²+n²+p²+q²+1\(\ge\) m(n+p+q+1) với mọi m, n, p, q

Answer 2

Giải:

Ta có:

\(m^2+n^2+p^2+q^2+1\ge m\left(n+p+q+1\right)\)

\(\Leftrightarrow\left(\dfrac{m^2}{4}-mn+n^2\right)+\left(\dfrac{m^2}{4}-mp+p^2\right)+\left(\dfrac{m^2}{4}-mq+q^2\right)+\left(\dfrac{m^2}{4}-m+1\right)\ge0\)

\(\Leftrightarrow\left(\dfrac{m}{2}-n\right)^2+\left(\dfrac{m}{2}-p\right)^2\) \(+\left(\dfrac{m}{2}-q\right)^2+\left(\dfrac{m}{2}-1\right)^2\) \(\ge0\) (luôn đúng)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{m}{2}-n=0\\\dfrac{m}{2}-p=0\\\dfrac{m}{2}-q=0\\\dfrac{m}{2}-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n=\dfrac{m}{2}\\p=\dfrac{m}{2}\\q=\dfrac{m}{2}\\m=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=2\\n=p=q=1\end{matrix}\right.\)

Vậy \(m^2+n^2+p^2+q^2+1\ge m\left(n+p+q+1\right)\) (Đpcm)