The equation $x^2+ax+(b+2)=0$ has real roots. Find the minimum value of $a^2+b^2$.

Question

The equation $x^2+ax+(b+2)=0$ has real roots. Find the minimum value of $a^2+b^2$.

Answer 1

We the discriminant $\ge0$ $$\implies a^2-4(b+2)\iff a^2\ge4(b+2)$$

$$a^2+b^2\ge4b+8+b^2=(b+2)^2+4\ge4$$

Answer 2

In general this type of question can be settled by the use of Cauchy - Schwarz inequality as follows: $x^2+2 = -ax - b \implies (x^2+2)^2 = (ax+b)^2 \leq (x^2+1)(a^2+b^2)\implies a^2+b^2 \geq \dfrac{(x^2+2)^2}{x^2+1} = \dfrac{((x^2+1) + 1)^2}{x^2+1}\geq 4$, by the well-known AM-GM inequality $(t+1)^2 \geq 4t, t = x^2+1 $ in this case.