\(\sqrt{\left(x+1995\right)^2}+\sqrt{\left(x+1996\right)^2}=\left|x+1995\right|+\left|x+1996\right|\)
\(=\left|-x-1995\right|+\left|x-1996\right|\)
Ta chứng minh Bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)
\(\Leftrightarrow\left(\left|a\right|+\left|b\right|\right)^2\ge\left(\left|a+b\right|\right)^2\)
\(\Leftrightarrow a^2+b^2+2\left|ab\right|\ge a^2+b^2+2ab\)
\(\Leftrightarrow\left|ab\right|\ge ab\) luôn đúng
Dấu = khi \(ab\ge0\)
\(\Rightarrow\left|-x-1995\right|+\left|x+1996\right|\ge\left|-x-1995+x+1996\right|=1\)
Dấu = khi \(\left(x+1995\right)\left(x+1996\right)\ge0\)\(\Rightarrow1995\le x\le1996\)
\(\Rightarrow\hept{\begin{cases}1995\le x\le1996\\\left(x+1995\right)\left(x+1996\right)=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1995\\x=-1996\end{cases}}\)
