\(\hept{\begin{cases}a+b+c+d=7\\a^2+b^2+c^2+d^2=13\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}b+c+d=7-a\left(1\right)\\b^2+c^2+d^2=13-a^2\left(2\right)\end{cases}}\)
Ta có:
\(\left(b+c+d\right)^2=b^2+c^2+d^2+2\left(bc+cd+db\right)\)
\(\le b^2+c^2+d^2+\left(b^2+c^2\right)+\left(c^2+d^2\right)+\left(d^2+b^2\right)=3\left(b^2+c^2+d^2\right)\)
\(\Rightarrow\left(b+c+d\right)^2\le3\left(b^2+c^2+d^2\right)\left(3\right)\)
Thế (1), (2) vào (3) ta được
\(\left(7-a\right)^2\le3\left(13-a^2\right)\)
\(\Leftrightarrow2a^2-7a+5\le0\)
\(\Leftrightarrow1\le a\le\frac{5}{2}\)
\(\Rightarrow\hept{\begin{cases}min\left(a\right)=1\\max\left(a\right)=\frac{5}{2}\end{cases}}\)
\(\Rightarrow\frac{min\left(a\right)+max\left(a\right)}{2}=\frac{1+\frac{5}{2}}{2}=\frac{7}{4}\)
