In order to make the original quadratic equation with one variable have two real roots, then

δ= -(2k+ 1)^2-4* 1*(k^2+2k)

= 4k^2+4k+ 1-4k^2-8k

= -4k+ 1

≥ 0

∴ k≤ 1/4

Therefore, the value range of k is k≤ 1/4.

(2)

From the relationship between the roots and coefficients of a quadratic equation (that is, Vieta's theorem)

x 1+x2= 2k+ 1，x 1x2= k^2+2k

x 1x2-x 1^2-x2^2= -(x 1+x2)^2+3x 1x2

= -(2k+ 1)^2+3(k^2+2k)

= -4k^2-4k- 1+3k^2+6k

= -k^2+2k- 1

= -(k- 1)^2

≥ 0

(k- 1)^2≥ 0

∴ (k- 1)^2= 0

∴ k= 1

But from the result of (1), when k≤ 1/4, the equation has two real roots, and k= 1 is not in this range.

So there is no such k, which makes the original inequality hold.

I hope you can adopt it, but you can ask if you don't understand it. thank you