\(P=\dfrac{a^2+b^2}{a-b}=\dfrac{\left(a-b\right)^2+2ab}{a-b}=\dfrac{\left(a-b\right)^2+2}{a-b}=\left(a-b\right)+\dfrac{2}{a-b}\)

Áp dụng bất đẳng thức Cauchy ta có:

\(\left(a-b\right)+\dfrac{2}{a-b}\ge2\sqrt{\left(a-b\right).\dfrac{2}{a-b}}=2\sqrt{2}\) hay \(P\ge2\sqrt{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a>b\\a-b=\dfrac{2}{a-b}\\ab=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=\sqrt{2}\\ab=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b+\sqrt{2}\\\left(b+\sqrt{2}\right)b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{\pm6+\sqrt{2}}{2}\\b=\dfrac{\pm\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)Vậy \(MinP=2\sqrt{2}\), đạt tại \(\left(a;b\right)=\left(\dfrac{\sqrt{6}+\sqrt{2}}{2};\dfrac{\sqrt{6}-\sqrt{2}}{2}\right),\left(\dfrac{-\sqrt{6}+\sqrt{2}}{2};\dfrac{-\sqrt{6}-\sqrt{2}}{2}\right)\)

a) Áp dụng bất đẳng thức Bnhiacopxki ta có :

\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(a.1+b.1+c.1\right)^2\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Rightarrow a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{3^2}{3}=3\)

b) Ta có : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đúng)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+ac+bc\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge3ab+3bc+3ac\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)

\(\Rightarrow ab+ac+bc\le\frac{\left(a+b+c\right)^2}{3}=\frac{3^2}{3}=3\)

\(A=2006-\frac{x}{6-x}\le2006\)

Min \(A=2006\Leftrightarrow\frac{x}{6-x}=0\Rightarrow x=0\)

\(B=\left|x-2001\right|+\left|x+1\right|\ge0\)

Min \(B=0\Leftrightarrow\hept{\begin{cases}x-2001=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2001\\x=-1\end{cases}}}\)