Given the base of a triangle, the opposite angle A, and the product `k^2` of the other two sides, show that it is not possible for `a` to be less than `2ksinA/2`

Explanation

Given `bc = k^(2)`

Now `cos A = (b^(2) + c^(2) - a^(2))/(2bc)`
or `2k^(2) cos A = b^(2)+ ((k^(2))/(b))^(2) -a^(2)`
or `b^(4) -(a^(2) + 2k^(2) cos A). B^(2) + k^(4) = 0`
Since `b^(2)` is real, `D ge 0 " or " (a^(2) + 2k^(2) cos A)^(2) - 4 k^(2) ge 0`
or `(a^(2) + 2k^(2) cos A + 2k^(2)) (a^(2) + 2k^(2) cos A - 2k^(2)) ge 0`
or `(a^(2) + 2k^(2) . 2 cos^(2).(A)/(2)) (a^(2) -2k^(2). 2 sin^(2).(A)/(2)) ge 0`
or `(a^(2) + 4k^(2) cos^(2). (A)/(2))(a^(2) - 4k^(2) sin^(2).(A)/(2)) ge 0`
or `a^(2) - 4k^(2) sin^(2).(A)/(2) ge 0`
[since `a^(2) (A)/(2) + 4k^(2) cos^(2) A` is always positive]
or `(a+ 2k sin. (A)/(2)) (a - 2 k sin.(A)/(2)) ge 0`
or `a le -2k sin.(A)/(2) " or " a ge 2k sin. (A)/(2)`
But a must be positive, which means `a le -2 k sin(A//2)` is rejected.
Hence, `a ge 2k sin. (A)/(2)`