MHB Find Minimum of Inequality Expression: 0<x<π/2

[Saitama]

I am trying to find the minimum of the following expression:
$$\frac{\sin^2x+8\cos^2x+8\cos x+\sin x}{\sin x\cos x}\,\,\,,0<x<\frac{\pi}{2}$$
I know I can bash this with calculus but the expression has a nice minimum value (=17) which makes me think that it can be solved by use of some inequality though I have no idea about how to proceed.

Any help is appreciated. Thanks!

 

[Opalg]

I am trying to find the minimum of the following expression:
$$\frac{\sin^2x+8\cos^2x+8\cos x+\sin x}{\sin x\cos x}\,\,\,,0<x<\frac{\pi}{2}$$
I know I can bash this with calculus but the expression has a nice minimum value (=17) which makes me think that it can be solved by use of some inequality though I have no idea about how to proceed.

Any help is appreciated. Thanks!

Let $t = \tan\frac x2$. Then $\sin x = \frac{2t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$. We want to show that $\sin^2x+8\cos^2x+8\cos x+\sin x \geqslant 17\sin x\cos x$, or $\sin x(\sin x+1) + 8\cos x(\cos x+1) - 17 \sin x\cos x \geqslant0.$ In terms of $t$ (after multiplying through by $(1+t^2)^2$), that becomes $$2t(2t+ 1+t^2) + 8(1-t^2)(1-t^2 + 1+t^2) - 34t(1-t^2) \geqslant0,$$ $$t(1+t)^2 + 8(1-t^2) - 17t(1-t^2) \geqslant0.$$ That simplifies to $18t^3 - 6t^2 - 16t + 8\geqslant0$, or $9t^3 - 3t^2 - 8t + 4\geqslant0$, which in turn factorises as $(3t-2)^2(t+1) \geqslant0.$ But $t>0$ because $x$ lies between $0$ and $\frac\pi2$. So that last inequality is evidently true (with equality holding only when $t = \frac23$), and you can work backwards to conclude that the original inequality holds for all $x$ in that interval.

The minimum of $\frac{\sin^2x+8\cos^2x+8\cos x+\sin x}{\sin x\cos x}$ occurs when $\tan\frac x2 = \frac23$, at which point $\sin x = \frac{12}{13}$ and $\cos x = \frac5{13}$. So the question is somehow based on the Pythagorean triple $(5,12,13)$.

 

[Saitama]

Let $t = \tan\frac x2$. Then $\sin x = \frac{2t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$. We want to show that $\sin^2x+8\cos^2x+8\cos x+\sin x \geqslant 17\sin x\cos x$, or $\sin x(\sin x+1) + 8\cos x(\cos x+1) - 17 \sin x\cos x \geqslant0.$ In terms of $t$ (after multiplying through by $(1+t^2)^2$), that becomes $$2t(2t+ 1+t^2) + 8(1-t^2)(1-t^2 + 1+t^2) - 34t(1-t^2) \geqslant0,$$ $$t(1+t)^2 + 8(1-t^2) - 17t(1-t^2) \geqslant0.$$ That simplifies to $18t^3 - 6t^2 - 16t + 8\geqslant0$, or $9t^3 - 3t^2 - 8t + 4\geqslant0$, which in turn factorises as $(3t-1)^2(t+1) \geqslant0.$ But $t>0$ because $x$ lies between $0$ and $\frac\pi2$. So that last inequality is evidently true (with equality holding only when $t = \frac23$), and you can work backwards to conclude that the original inequality holds for all $x$ in that interval.

The minimum of $\frac{\sin^2x+8\cos^2x+8\cos x+\sin x}{\sin x\cos x}$ occurs when $\tan\frac x2 = \frac23$, at which point $\sin x = \frac{12}{13}$ and $\cos x = \frac5{13}$. So the question is somehow based on the Pythagorean triple $(5,12,13)$.

This is nice! Thanks a lot Opalg! (Sun) (Bow)