XY + YZ + ZX = 1, find x √ YZ + y √ ZX + Z √ XY Is less than or equal to

XY + YZ + ZX = 1, find x √ YZ + y √ ZX + Z √ XY Is less than or equal to

This paper examines the inequality of maximum value
A + B ≥ 2 √ AB if and only if a = B, take the equal sign
x√yz+y√zx+z√xy
≤x(y+z)/2+y(z+x)/2+z(x+y)/2
If and only if y = Z, z = x, x = y, that is, x = y = Z, we take the equal sign
x√yz+y√zx+z√xy
≤(xy+xz+yz+xy+xz+yz)/2
=xy+yz+xz
=1
Therefore:
X √ YZ + y √ ZX + Z √ XY ≤ 1, when x = y = Z, take the equal sign