Prove that any two sides of a triangle are together greater than twice the median drawn to the third side. GIVEN : ` A B C` in which `A D` is a median. PROVE : `A B+A C >2A D` CONSTRUCTION : Produce `A D` to `E` such that `A D=D Edot` Join `E Cdot`

Explanation

`27a^3 + (1/646)^3+27a^2/4b +9a/166^2`...(i)
`(3a)^3+ (1/4b)^3 + 3(3a) (3a)(1/4b) + 3(3a)(1/4b)(1/4b)`...(ii)
where `a = 3a and b = 1/4b`
then, `(3a+1/4b)^3`are the factors of `27a^3 + 1/64b^3 + (27a^2)/4b+(9a)/166^2`.
Hence, `27a^3 + 1/64b^3 + (27a^2)/4b+(9a)/166^2` can be factorized as `(3a+1/4b)^3`.