√sinx+√cosx>=1 how?? in [0,pi/2}?

Squaring
SinX + 2√SinxCosx + CosX = 1
2√SinxCosx = 1 - SinX - CosX
Square up again.
4SinxCosx = ( 1 - SinX - CosX)^2
2Sin(2x) = ( 1 - (SinX + CosX))^2
2Sin(2x) = 1 - 2(SinX + CosX))+ (SinX + CosX)^2
2Sin(2x) = 1 - 2(Sinx + CosX) + Sin^2X + Cos^2X + 2(SinX + CosX)
2Sin(2x) = 1 + Sin^2X + Cos^2X
2Sin(2x) = 1 + 1 = 2
Sin(2x) = 2/2 = 1
Sin(2x) = 1
2x = Sin^-1(1) = pi/2 (90 degrees)
x = pi/4 (45 degrees)

Sin45 = Cos45 = 1/sqrt(2) = sqrt(2) / 2

Yes!!! There is some 'heavy' algebra in it. First you have to remove the 'sqrt's', by 'double' squaring.
Then apply 'quadratic' expansion.
Finally remember the Trig Identities (S^2 + C^2 = 1 )& 2SC = SC + CS = S(2)

Hope that helps!!!!

0 < sin x < 1
0 < cos x < 1

observe that 1/2 = sqrt(1/4) and 1/2 > 1/4

now does this makes sense