Notice that 2x² + 2x > -4 means that :

2x² + 2x + 4 > 0

So, if you can prove this, you would be able to prove the original inequality.



We can start by completing the square for 2x² + 2x + 4.


2x² + 2x + 4

= 2(x² + x + 2)

= 2(x² + 2x(½) + (½)² - (½)² + 2)

= 2((x + ½)² + 7/4)

= 2(x + ½)² + 7/2

Now, (x + ½)² ≥ 0 for all real values of x since the square of any real value is either 0 or positive.

Eg. 5² = 25 , 0² = 0, (-9)² = 81

So regardless of whether x + ½ is positive, negative, or 0, (x + ½)² will never be negative for all real values of x.


Then, continuing from above,

For all real values of x,

2(x + ½)² ≥ 0
(multiplying by a positive number 2 means that it is still 0 or positive. So no change in the sign direction)

2(x + ½)² + 7/2 ≥ 7/2

2x² + 2x + 4 ≥ 7/2
(Recall that we had shown that 2x² + 2x + 4 = 2(x + ½)² + 7/2 . Alternatively you can think of it as expanding the expression)

2x² + 2x + 4 > 0

(Since 2x² + 4x + 4 ≥ 7/2 , it is never less than 7/2. Which means that it is always bigger than 0 since 7/2 is greater than 0)


2x² + 2x > -4

(Proved)

(Edited for typos at 27 Jan 2021 2.18pm)

But in doing so, we are technically assuming that the statement 2x2 + 2x + 4 > 0 is already true.

2x2 + 2x
= 2 (x2 + x)
= 2 (x2 + x + 0.5^2 - 0.5^2)
= 2 [(x + 0.5^2) - 0.25]
= 2 (x + 0.5)^2 - 0.5
>= -0.5 for all x
> -4 for all x

Not quite.

If you assume something to be true, you are already using the result from the start.

i.e ① 2x² + 2x > -4 would have been the first line, and ② 2x² + 2x + 4 > 0 would be used after.


Here, the idea is to first show that ② 2x² + 2x + 4 > 0 by first showing ③ 2x² + 2x + 4 ≥ 7/2. And this in turn is shown by using ④ 2(x + ½)² ≥ 0

Then, since ④ →③ → ②, which then leads to ①.

In the working to be presented,, there was never any assumption that ① or ② was true. In fact, it is working towards ① by a series of workings.

In short : show that ② is true first, then th at implies that ① is true .

If you had actually noticed, your working is actually analogous to what was written above.

The difference was that while you started with the LHS right away, I chose to use 2x² + 2x + 4. And mine has an additional step at the end.

If written in a similar fashion to yours, it would simply be :


2x² + 2x + 4

= 2(x² + x + 2)

= 2(x² + 2x(½) + (½)² - (½)² + 2)

= 2((x + ½)² + 7/4)

= 2(x + ½)² + 7/2

≥ 7/2 for all real x

> 0 for all real x

Then, since 2x² + 2x + 4 > 0 for all real x,

2x² + 2x > -4 for all real x.