# 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.

The question is equivalent to $$2 d (a-b+c)+2 (a-b) (b-c)-2 d^2\geq 0.$$ The second term term is obviously $\geq 0$. Meanwhile, the first term and the third term can be factored as $2 d (a-b+c-d)$, which is also $\geq 0$.

Thus the entire expression is $\geq 0$.

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.
• How did you get the first line? I can't seem to see it... Commented Mar 31, 2014 at 16:04
• @YiyuanLee ? First line is just added parentheses to the original problem Commented Mar 31, 2014 at 16:04
• + 1 Now that was embarrassing for me >< Commented Mar 31, 2014 at 16:05

A standard trick for a problem like this is to make a variable substitution which simplifies the problem in some way; hopefully you hit on a substitution which makes the problem more amenable. One idea for such a substitution is to simplify the condition that $a\geq b\geq c\geq d\geq 0$, and we can do this by defining new variables representing the difference between consecutive original variables: \begin{align} x &= a - b \\ y &= b - c \\ z &= c - d. \end{align} Using the new variables, we find that the condition becomes $x,y,z,d\geq0$. Next, replace $a$, $b$, and $c$ in the inequality to be proved (we're working backwards here). The lefthand side becomes \begin{align} a^2 - b^2 + c^2 - d^2 &= (a-b)(a+b) + (c-d)(c+d) \\ &= x\cdot [(x+y+z+d) + (y+z+d)] + z\cdot [(z + d) + d] \\ &= x^2 + 2xy + 2xz + 2xd + z^2 + 2zd \\ &= (x^2 + 2xz + z^2) + 2xy + 2xd + 2zd. \end{align} The righthand side becomes $$(a-b + c-d)^2 = (x + z)^2 = x^2 + 2xz + z^2.$$ With the new variables, your inequality is obvious! In particular, we see that equality holds if and only if $xy + xd + zd = 0$, that is, if and only if one of the following is true:

• $x=0$ and either $z=0$ or $d=0$
• $y=0$ and $d=0$.

In other words, these conditions say that

• $a=b$ and either $c=d$ or $d=0$
• $b=c$ and $d=0$.