parmenides51 wrote: Let $a$ and $b$ be real numbers, where $a \not= 0$ and $a \not= b$ and all the roots of the equation $ax^{3}-x^{2}+bx-1 = 0$ is a real and positive number. Find the smallest possible value of $P = \dfrac{5a^{2}-3ab+2}{a^{2}(b-a)}$. (Heir of Ramanujan) P5

parmenides51 wrote: Let $a$ and $b$ be real numbers, where $a \not= 0$ and $a \not= b$ and all the roots of the equation $ax^{3}-x^{2}+bx-1 = 0$ is a real and positive number. Find the smallest possible value of $P = \dfrac{5a^{2}-3ab+2}{a^{2}(b-a)}$. Ugly algebraic approach : Note that sum of roots (all positive) is $\frac 1a$ and so $a>0$ We can WLOG consider equation $P(\frac xa)=0$, which is $x^3-x^2+ux-v=0$ where $u=ab$ and $v=a^2$, with $u,v>0$ and $u\ne v$ And we have to minimize $P=\frac{5v-3u+2}{(u-v)\sqrt v}$ Let $r_1,r_2,r_3$ be the three positive roots. We have $1+\frac uv=\sum r_i+\sum\frac 1{r_i}\ge 6$ (since each $r_i+\frac 1{r_i}\ge 2$) and so $\frac uv\ge 5$ Writing equation as $x^2-x^3=ux-v$, we have tangent at inflexion point $y=\frac x3-\frac 1{27}$ and so, in order to have three roots, we need $u\le\frac 13$ and $v\le \frac 1{27}$ And so $\frac 13\ge u\ge 5v\ge \frac 5{27}$ So $u>v$ and $5v-3u+2\ge 5v-1+2>0$ and $P>0$ So, for a given $v$, $P$ is a decreasing function of $u$ and so $u$ is the slope of tangent to second part of $x^2-x^3$ (over $[\frac 13,\frac 23)$ passing thru point $(0,-v)$ Let $t\in[\frac 13,\frac 23)$ the tangency point. We have $u=2t-3t^2$ and $v=t^2-2t^3$ (and so $t\le\frac 12$ in order $v>0$) And so $P=\frac{5t^2-2t+1}{t^2(1-t)\sqrt{1-2t}}$ and we are looking for $\min_{t\in[\frac 13,\frac 12)}\frac{5t^2-2t+1}{t^2(1-t)\sqrt{1-2t}}$ This function is increasing over given interval and minimum is reached for $t=\frac 13$ and is $\boxed{12\sqrt 3}$ (with original equation $\frac{\sqrt 3}9x^3-x^2+\sqrt 3x-1=(\frac x{\sqrt 3}-1)^3$)