Ta có: \(\left|x-8\right|+\left|6-x\right|\ge\left|x-8+6-x\right|=2\forall x\)
\(\left(y+3\right)^2\ge0\forall y\)
=>\(5\left(y+3\right)^2\ge0\forall y\)
=>\(5\left(y+3\right)^2+12\ge12\forall y\)
=>\(\frac{24}{5\left(y+3\right)^2+12}\le\frac{24}{12}=2\forall y\)
Ta có: \(\left|x-8\right|+\left|6-x\right|=\frac{24}{5\left(y+3\right)^2+12}\)
mà \(\left|x-8\right|+\left|6-x\right|\ge\left|x-8+6-x\right|=2\forall x\)
và \(\frac{24}{5\left(y+3\right)^2+12}\le2\forall y\)
nên dấu '=' xảy ra khi \(\begin{cases}\left(x-8\right)\left(6-x\right)\ge0\\ y+3=0\end{cases}\Rightarrow\begin{cases}\left(x-8\right)\left(x-6\right)\le0\\ y=-3\end{cases}\Rightarrow\begin{cases}6\le x\le8\\ y=-3\end{cases}\)
