Triangle Inequality.

[man111]

If [tex]0<x<\frac{\pi}{4}[/tex].Then prove that [tex]\frac{cosx}{sin^2x.(cosx-sinx)}>8[/tex]

Here is my process........

Let [tex]A=\frac{cosec^2x}{1-tanx}=\frac{1+cot^2x}{1-tanx}=\frac{t^2+1}{t^2.(1-t)}[/tex] (Where [tex]t=tanx[/tex] and [tex]x\in (0,1)[/tex])

My question is how can i proceed from here .........

Thanks.

[Mark]

[tex]\frac{8t^3-7t^2+1}{t^2(1-t)}>0[/tex] and [tex]t\in (0,1)[/tex] thus we need to prove [tex]8t^3-7t^2+1>0.[/tex]Let[tex]f(t)=8t^3-7t^2+1, t\in (0,1)[/tex][tex]f'=24t^2-14t, f'=0, t=\frac{7}{12},[/tex] [tex]f''=48t-14,[/tex] [tex]f''(\frac{7}{12})>0[/tex] therefore [tex]minf(t)=f(\frac{7}{12})=\frac{89}{432}>0[/tex]