Real numbers/x^2+(x+1)^2 \geq (x+2)^2 for x \geq 3/Exercise/Solution

We compute both sides, the estimate we have to show is then

${}2x^{2}+2x+1\geq x^{2}+4x+4\,.$

For real numbers we can add on both sides, so the estimate is equivalent to

${}x^{2}-2x-3\geq 0\,.$

We write the left hand side as

${}x^{2}-2x-3=x^{2}-2x+1-1-3=(x-1)^{2}-4\,.$

For ${}x\geq 3$ we have ${}x-1\geq 2$ and therefore

${}(x-1)^{2}\geq 2(x-1)\geq 2^{2}=4\,.$

So for ${}x\geq 3$ we get the estimate

${}(x-1)^{2}-4\geq 0\,$

and hence the original estimate.

To the exercise