1. ## Inequality holds

Hi, there's a proof I need to complete which requires that the following inequality holds:

(1/4)*((b-a[1])^2+(b-a[2])^2+(1-a[3])^2+(1-a[4])^2) >= 1/6*((b-a[1])^2+(b-a[2])(2*b-a[1]-a[2])+(b-a[3])(2*b-a[4]-a[3])+(b-a[4])^2)

where:

0<=a[1]<=a[2]<=a[3]<=a[4]<=b<=1

I have made several plots and in all of them the inequality holds. Nonetheless, I need to finish this problem theoretically. I have already made the proof for the case of a[1]=a[2] and a[3]=a[4], and also for the case where a[1]=a[2]=a[3]=a[4], but these are straightforward, and the real deal comes when I want to solve for the general case.

I know 0<=(2*b-a[4]-a[3])<=(2*b-a[1]-a[2])<=2. I also know that (b-a[3])(2*b-a[4]-a[3]) <= (b-a[2])(2*b-a[1]-a[2]). Finally, I also know that the case where both sides of the inequality are maximums occurs when a[1]=a[2]=a[3]=a[4]=0 and b=1.

I hope somebody here can help me with this. Best regards.

2. I'll repost the inequality using LaTeX code:
$\frac{1}{4}((b-a_1)^2+(b-a_2)^2+(1-a_3)^2+(1-a_4)^2) \geq$
$\frac{1}{6}((b-a_1)^2+(b-a_2)(2\cdot b-a_1-a_2)+(b-a_3)(2\cdot b-a_3-a_4)+(b-a_4)^2)$

I hope it will be better this way.

3. Is this the right place in the forum to ask for this kind of questions? Or should a move it? I mean, 18 views and not a single idea? Please, let me know as I really need to solve it. Thanks in advance.