if tan^-1x + tan^-1y +tan^-1z = π/2 And x+y+z = Root 3 then prove that x=y=z

x+y+z=sqrt(3)Squaring gives you x^2 + y^2 + z^2 + 2(xy+yz+zx) = 3Now I will use atan(x) to refer to tan inverse of x.If atan(x)+atan(y)+atan(z) = pi/2, then atan(x)+atan(y) = pi/2 – atan(z) = acot(z) = atan(1/z)So simplifying the lhs there, we get atan(x+y / 1-xy) = atan(1/z)Expanding and simplifying gives us xy+yz+zx=1We can substitute this in the first relation we got! Giving us x^2 + y^2 + z^2 = 1.Now, similar to AM≥GM, we also have QM≥AM, where QM is the ‘quadratic mean’, but we commonly refer to it as ‘root mean square’. So, using this inequality, we can say that sqrt(x^2+y^2+z^2 / 3) ≥ x+y+z / 3But using the information we have, this gives us 1/sqrt(3) ≥ 1/sqrt(3)Now, in QM≥AM, equality holds if and only if x=y=z, just like with AM≥GM!So we must have x=y=z !Side note: We can keep going on with these inequalities!We can define the nth power mean of two numbers as (a^n + b^n / n) ^ 1/n. You can see that the 2nd power mean is the root mean square, and the first power mean is the arithmetic mean. You can even see how the 0th power mean is the geometric mean, as a limiting case!And one important thing here is that the nth power mean of any numbers is an increasing function of n! Its value is bigger if n gets bigger. This is why QM≥AM≥GM: because 2>1>0 !Another fun fact: for n tending to +infinity, the nth power mean of a and b becomes max(a, b). And for n tending to -infinity, it becomes min(a, b)!