Question

If
`x` is real and p=
`(3(x^(2) +1))/(2x-1)`, prove that
`p^(2)`-3(p+3) ≥ 0

Answer 1

`{ \tt{p=(3(x^(2) +1))/(2x-1)}} \\ { \tt{p(2x - 1) = 3( {x}^(2) + 1) }} \\ { \tt{2px - p = 3 {x}^(2) + 3 }} \\ { \tt{3 {x}^(2) - (2p)x + (p + 3) = 0}}`

By factorization :

`{ \tt{ ( {p}^(2) - 3)( p + 3) \: is \: the \: zero}}`

Since the roots are real, they're greater than zero ( 0 < x ≤ +∞ ):

`{ \tt{ ({p}^(2) - 3})(p + 3) \geqslant 0}`