Minimum value of $(x+1)^2+3$ with AM-GM

Question

Find the minimum value of $$(x+1)^2+3$$

I don't know how to use the AM-GM inequality.

I think in this case $a = (x+1)^2$ and $b = 3$. So, $$(x+1)^2+3 \geq 2\sqrt{3(x+1)^2}$$

And from here I don't know what I should do.

Answer 1

As noted already (deleted since), the inequality does not lend itself to a natural proof by AM-GM.

For an overstrained proof, using just $\color{red}{\text{AM-GM}}$ and $\sqrt{x^2}=\color{blue}{|x| \ge -x}\,$:

$$(x+1)^2+3 = \color{red}{x^2+1} + 2x + 3 \ge \color{red}{2|x|} + 2x + 3 = 2(\color{blue}{|x|+x})+3 \ge \color{blue}{0} + 3 = 3$$

The lower bound of $3$ is attained for $x=-1\,$, so it is in fact a minimum.

Answer 2

Why you used A.M-G.M. inequality. It can be done more easily.

Answer: We know that every square number is nonnegative. So $(x+1)^2\le 0$. Therefore, $(x+1)^2+3\le 3$. Finally, the lower bound is 3 which attained when $(x+1)^2=0$, or, $x=-1$.