Four Variables, One Equation?

$a^2+b^2+c^2+1=d+\sqrt{a+b+c-d}$

add $a+b+c$ to both sides and take $d$ and $\sqrt{a+b+c-d}$ to the left

$a^2+b^2+c^2+(a+b+c-d)-\sqrt{a+b+c-d}+1=a+b+c$

now take $\frac{1}{4}$ out of the $1$ and we can rewrite the equation as

$a^2+b^2+c^2+(\sqrt{a+b+c-d}-\frac{1}{2})^2+$ + $\frac{3}{4}$ = $a+b+c$

on further rearrangement we get

$(a^2-a+\frac{1}{4})+(b^2-b+\frac{1}{4})+(c^2-c+\frac{1}{4})+(\sqrt{a+b+c-d}-\frac{1}{2})^2=0$

or

$(a-\frac{1}{2})^2+(b-\frac{1}{2})^2+(c-\frac{1}{2})^2+(\sqrt{a+b+c-d}-\frac{1}{2})^2=0$

$\Rightarrow$ $a=b=c=\frac{1}{2}$ and $d=\frac{5}{4}$

Writing this as a quadratic equation in $d$ gives the discriminant $4a-4a^2+4b-4b^2+4c-4c^2-3$ . Now the maximal value of $x-x^2$ is $\frac{1}{4}$ , attained at $x=\frac{1}{2}$ . Thus the discriminant is nonnegative only when $a=b=c=\frac{1}{2}$ , in which case we find $d=\boxed{\frac{5}{4}}$

Conversion obtained:

d = a^2 + b^2 + c^2 + 1/2 +/- Sqrt (a + b+ c - a^2 - b^2 - c^2 - 3/4)

Since not easy to get real d with discriminant < 0 usually, let x = a = b = c:

Discriminant = - 3 x^2 + 3 x - 3/4 = -3 (x - 1/2)^2

With x = 1/2 such that a = b = c = 1/2, Discriminant = 0 as an only hope.

d = 3 (1/2)^2 + 1/2 + 0 = 1.25

$a^2 + b^2 + c^2 + 1 = d + \sqrt{a+b+c-d}$

I thought that writing $d$ as " $a+b+c$ + something " would make the square root much friendlier. Then I thought, writing it as " $a+b+c$ - something " would make it even more friendlier. So :

Let $d = a+b+c - f(a,b,c,d)$

I then noticed that substituting the other $d$ as well would lead to a very nice surprise. Indeed:

$a^2 + b^2 + c^2 + 1 = a+b+c - f(a,b,c,d) + \sqrt{f(a,b,c,d)}$

can be rewritten as :

$(f(a,b,c,d) - \sqrt{f(a,b,c,d)} + \frac1 4) + (a^2- a + \frac 1 4) + (b^2- b + \frac 1 4) + (c^2- c + \frac 1 4)=0$

In other words :

$(\sqrt{f(a,b,c,d)}-\frac 1 2)^2 + (a- \frac 1 2)^2 + (b - \frac 1 2)^2 + (c- \frac 1 2)^2 = 0$ (constantly hunting for factorisations made it easier to see this idea)

Thus we obtain : $\sqrt{f(a,b,c,d)} = a = b = c = \frac 1 2$ .

And finally $d = \frac 5 4 = 1.25$

Let C1 = a^2 + b^2 + c^2 + 1, C2 = a + b + c, u = sqrt(C2 - d) Equation will become system: C1 - d = u and C2 - d = u^2 => u^2 - u = C2 - C1 Delta of above equation is del = 1 + 4C2 - C1 = -(2a - 1)^2 - (2b-1)^2 - (2c-1)^2 <=0 Therefore, a = b = c = 1/2 => C1 = 7/4 Moreover, u will have only 1 root and it's u = 1/2 Thus, d = C1 - u = 5/4

a=b=c=1/2,Therefore L.H.S= a+b+c+1=1/4+1/4+1/4+4/4 =7/4 and R.H.S = 5/4+ root of2/4+2/4+2/4-5/4=5/4+root of 1/4 ie 5/4+1/2=5/4+2/4 = 7/4 Hence LHS=RHS= 1.25