Question

chứng minh rằng\(\dfrac{a^2+a+2}{\sqrt{a^2+a+1}}>=2\) với a thuộc i

Answer 1

\(\dfrac{a^2+a+2}{\sqrt{a^2+a+1}}\ge2\Leftrightarrow a^2+a+2\ge2\sqrt{a^2+a+1}\Leftrightarrow\left(a^2+a+1\right)-2\sqrt{a^2+a+1}+1\Leftrightarrow\left(\sqrt{a^2+a+1}-1\right)^2\ge0\)

Answer 2

\(\dfrac{a^2+a+2}{\sqrt{a^2+a+1}}=\dfrac{a^2+a+1}{\sqrt{a^2+a+1}}+\dfrac{1}{\sqrt{a^2+a+1}}=\sqrt{a^2+a+1}+\dfrac{1}{\sqrt{a^2+a+1}}\ge2\)

Dấu "=" xảy ra <=> a² + a + 1 = 1

<=> a(a + 1) = 0

<=> a = 0 hoặc a = -1