Solution matters more than the answer!

Solution: Notice that $a+ 1 \geq \frac{(2a-1)^2}{a^2 - a + 1} \iff a^3 + 1 \geq 4a^2 - 4a + 1 \iff a^3 - 4a^2 + 4a \geq 0 \iff a(a-2)^2 \geq 0$ , which is true.

Thus, $\frac{(2a-1)^2}{a^2 - a + 1} + \frac{(2b-1)^2}{b^2 - b + 1} + \frac{(2c-1)^2}{c^2 - c + 1} + \frac{(2d-1)^2}{d^2 - d + 1} \leq (a+1) + (b+1) + (c+1) + (d+1) = \boxed{8}$ .

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Wait, where did that first inequality come from???

The answer to this question is a trick called the "tangent line trick". The tangent line trick is a trick which doesn't work so well in finding the maximum/minimum values of expressions, but it does great when it comes to proving inequalities.

The first step is to randomly guess the possible tuple $(a,b,c,d)$ which yields the maximum value. A little bit of experimentation should yield the maximum to be $8$ , after trying out cases when $a = b = c = d$ , $a = 0$ , $a = b = 0$ and $a = 4$ .

Now that we've gotten the maximum $(8)$ and the equality case $(a = b = 2, c = d = 0)$ , we just need to prove this is the maximum. Here, we use the trick.

Notice that our inequality is of the form $f(a) + f(b) + f(c) + f(d) \leq 8$ with $a + b + c + d = 4$ and $f(x) = \frac{(2x-1)^2}{x^2 -x + 1}$ . The trick here is to take the line tangent to $f(x)$ at the equality case (when $x = 2$ ) and prove that this line is greater than or equal to $f(x)$ .

(Note that this works precisely because our constraint is of the form $a + b + c + d = k$ , and if we add up linear equations, we're bound to get the expression $a+b+c+d$ somewhere in the sum, allowing us to use our constraint.)

Now, note that the tangent line to $f(x)$ at $x=a$ has the form $f(a) + f'(a)(x-a)$ . A little differentiation shows us that $f'(2) = 1$ and we have that $f(2) = 3$ , giving us the tangent line to $f(x)$ at $x = 2$ to be $3 + 1(x-2) = x+1$ . Thus, our possible inequality becomes

$x+ 1 \geq \frac{(2x-1)^2}{x^2 - x + 1}$

Which we proved above with equality holding when $x = 0$ or $x = 2$ . Now, we proceed as we did above.