Question

Chứng minh rằng:

\(\frac{a^2+a+2}{\sqrt{a^2+a+1}}\ge2\) với mọi a.

Answer 1

\(a^2+a+1=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall a\)

\(P=\frac{a^2+a+1+1}{\sqrt{a^2+a+1}}=\sqrt{a^2+a+1}+\frac{1}{\sqrt{a^2+a+1}}\ge2\) (Cô-si)

Dấu "=" xảy ra khi \(a^2+a+1=1\Rightarrow\left[{}\begin{matrix}a=0\\a=-1\end{matrix}\right.\)