Is this sum lesser or equal to 2? Given any $a, b, c, d \in [0, 1],$ how can the following be proven?
$$a (1-b) + b(1-a) + c(1-d) + d(1-c) \le 2$$
 A: (To those who originally upvoted: I've changed my answer because this is a much sleeker version, not sure why I didn't see it before.)
This is just a single inequality added to itself with $c,d$ instead of $a,b$ in the second copy. Namely,
$$a(1-b)+(1-a)b\le a(1)+(1-a)(1)\le1.$$
Above we used the fact that if $b\in[0,1]$, then $b\le1$ and $1-b\le1$.
A: I agree with anon's solution. I'd also like to post mine. Anon's solution is based on a "trick" with inequalities, which is understood, but not so simple to guess from the beginning. OTOH my solution is straightforward.
$$S = a(1−b)+(1−a)b$$
Let's find the derivative of the above with respect to $a$.
$$\frac{dS}{da} = 1-2b.$$
From the above we see that the derivative of $S$ with respect to $a$ is constant (i.e. does not depend on $a$). Hence the extremal values are achieved at the ends of the range.


*

*Assuming $1-2b > 0, b < 1/2$. The maximum is achieved when $a=1$. Substituting this we get $S = 1-b$. Since $b\ge0$ we get $S \le 1$.

*Assuming $1-2b < 0, b > 1/2$. The maximum is achieved when $a=0$. Substituting this we get $S = b$. Since $b<=1$ we get $S \le 1$.

*If $1-2b = 0$, then $b = 1/2, S = 1/2 \le 1$.

