Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) ta có:
\(VT=\sqrt{10-x}+\sqrt{x+6}\)
\(\ge\sqrt{10-x+x+6}=\sqrt{16}=4\)
Và \(VP=-x^2-12x-32=-x^2-12x-36+4\)
\(=-\left(x^2+12x+36\right)+4\)\(=-\left(x+6\right)^2+4\le4\)
\(\Rightarrow VT\ge VP=4\) xảy ra khi \(VT=VP=4\)
\(\Rightarrow-\left(x+6\right)^2+4=4\Rightarrow x=-6\)