<s>[INST] Prove that $a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2$ In thinking about a base case in this problem, I came up with the following question.
Given real numbers $a \ge b \ge c \ge d \ge 0$, prove that the following holds:
$a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2 \tag{A}$
My attempt:
After simplification, this reduces to proving the inequality:
$\underbrace{ab}_{(1)} + \underbrace{bc}_{(2)} + \underbrace{cd + da}_{(3)}  \ge  \underbrace{ac}_{(1)} + \underbrace{bd}_{(2)} + \underbrace{b^2 + d^2}_{(3)} \tag{B}$
I tried to attack pairs of terms individually. This gave pairs  $(1)$ and $(2)$ which satisfied the $\ge$ relation, since $ab \ge ac \implies b \ge c$ true and $bc \ge bd \implies c \ge d$ true.
But then I got stuck at proving pair $(3)$ also satisfied the $\ge$ relation. That is $cd + da \ge b^2 + d^2 \tag{C}$
It turned out that $(C)$ doesn't hold in general. For example $(a, b, c, d) = (5, 4, 3, 2)$ gives $3 \cdot 2 + 2 \cdot 5 \ge 4^2 + 2^2 \implies 16 \ge 20$ false.
So, my strategy was incorrect. I would appreciate if anyone could show me the right approach for proving either (A) or (B).
Update: See this for a generalization of this problem.
 [/INST] Here's a method without expanding everything:
$$\begin{align}(a^2-b^2)+(c^2-d^2)&\ge((a-b)+(c-d))^2\\
(a-b)(a+b)+(c-d)(c+d)&\ge(a-b)^2+(c-d)^2+2(a-b)(c-d)\\
(a-b)(a+b-(a-b))+(c-d)(c+d-(c-d)&\ge2(a-b)(c-d)\\
(a-b)b+(c-d)d&\ge(a-b)(c-d)\\
(a-b)(b-c+d)+(c-d)d&\ge0\end{align}$$
And that's obvious. If you want to see when equality holds, consider all $4$ cases:


*

*$a=b$ and $c=d$

*$a=b$ and $d=0$

*$b=c-d$ and $d=0$, that is $b=c$ and $d=0$

*$b=c-d$ and $c=d$, this implies $b=c=d=0$ and is already handled by the previous case.

</s>