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

$2x^2+2x+1\ge x^2+4x+4.$

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

$x^2-2x-3\ge 0.$

We write the left hand side as

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

For $x\ge 3$ we have $x-1\ge 2$ and therefore

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

So for $x\ge 3$ we get the estimate

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

and hence the original estimate.

To the exercise