Áp dụng bất đẳng thức Cauchy ta có:
\(m^2+n^2+p^2+q^2+1\)
\(=\left(\frac{1}{4}m^2+n^2\right)+\left(\frac{1}{4}m^2+p^2\right)+\left(\frac{1}{4}m^2+q^2\right)+\left(\frac{1}{4}m^2+1\right)\)
\(\ge2\sqrt{\frac{1}{4}m^2\cdot n^2}+2\sqrt{\frac{1}{4}m^2\cdot p^2}+2\sqrt{\frac{1}{4}m^2\cdot q^2}+2\sqrt{\frac{1}{4}m^2\cdot1}\)
\(=2\cdot\frac{1}{2}mn+2\cdot\frac{1}{2}mp+2\cdot\frac{1}{2}mq+2\cdot\frac{1}{2}m\)
\(=mn+mp+mq+m\)
\(=m\left(n+p+q+1\right)\)
Dấu "=" xảy ra khi: \(\frac{1}{4}m^2=n^2=p^2=q^2=1\)\(\Rightarrow\hept{\begin{cases}m=2\\n=p=q=1\end{cases}}\)
