If sin^2 A/2 +sin^2 B/2 + sin^2 C/2 = 1-2 sinA/2 sinB/2 sinC/2 then prove that sin^2 A/2 + sin^2 B/2 +sin^2 C/2 >= 3/4

Sin^2A/2+sin^2B/2+sin^2c/2=1-2sinA/2sinB/2sinC/2.if and only if A+B+C=180° In ∆ cosA+cosB+cosC=1+4sinA/2sinB/2sinC/2max value of cosA+cosB+cosC=3/2 (I had already posted proof for this)So by substituting we get sinA/2sinB/2sinC/2=1/8So 1-2sinA/2sinB/2sinC/2=1-2×1/8=3/4Therefore min value of sin^2A/2+sin^2B/2+sin^2C/2=3/4