I have two polynomials, namely, $$f(y)=y^3-(k+3)y^2+2(k+1)y-2,$$ $$g(y)=y^2-(k+2)y+2.$$ where $k\ge4$ is a natural number. I want to prove that smallest root of $f$ is strictly less than smallest root of $g$.
I thought of calculating roots of $f$ and $g$ explicitly but that calculation is not very easy for $f$.
Next approach is: I already know that all roots of $f$ and $g$ are positive. We have smallest root of $g$ as $\lambda=\frac{(k+2)-\sqrt{(k+2)^2-8}}{2}$. We know that $f(0)<0$. So if we can prove that $f(y)>0$ for some $y \in (0,\lambda]$. Then we are done. Now
$f(y)=y(g(y)+2k-y)-2.$
So, $f(\lambda)=\lambda(2k-\lambda)-2$. But I am not sure that $f(\lambda)>0$.
How to prove the result?
Note that $$f(y) = (y-1)g(y) + (k-2)y$$ so if $g(y)=0$ then $f(y) = (k-2)y$. Since for $k>2$ both roots of $g$ are positive it follows that $f(y) > 0$.