Differentiation

Part (i).

To locate the stationary point of $y=\mathrm{f}(\beta )=\beta -\frac{1}{\beta }-\frac{1}{{\beta }^2}$, differentiate and set equal to zero:$${\mathrm{f}}^{\prime }(\beta )=1+\frac{1}{{\beta }^2}+\frac{2}{{\beta }^3}=\frac{{\beta }^3+\beta +2}{{\beta }^3}.$$The numerator factors as $(\beta +1)({\beta }^2-\beta +2)$. The quadratic factor has discriminant $1-8=-7<0$, so it is always positive. The only real root is therefore $\beta =-1$, giving the unique stationary point $$\begin{gathered} (-1, \\ ,\mathrm{f}(-1))=(-1, \\ ,-1) \end{gathered}$$.

Since ${\mathrm{f}}^{\prime }(\beta )>0$ for $\beta >0$ (all three terms are positive) and the single stationary value occurs at $\beta =-1$ with ${\mathrm{f}}^{\prime \prime }(-1)=-4<0$, the point $(-1,-1)$ is a local maximum on the branch $\beta <0$. The curve has a vertical asymptote at $\beta =0$ and behaves like $y\sim \beta$ for large $|\beta |$. Both branches tend to $-\infty$ as $\beta \to 0$.

For $y=\mathrm{g}(\beta )=\beta +\frac{3}{\beta }-\frac{1}{{\beta }^2}$, differentiate:$${\mathrm{g}}^{\prime }(\beta )=1-\frac{3}{{\beta }^2}+\frac{2}{{\beta }^3}=\frac{{\beta }^3-3\beta +2}{{\beta }^3}=\frac{(\beta -1{)}^2(\beta +2)}{{\beta }^3}.$$The roots are $\beta =1$ (double) and $\beta =-2$. Since $(\beta -1{)}^2⩾0$ always, the sign of ${\mathrm{g}}^{\prime }$ on $\beta <0$ is determined by $(\beta +2)/{\beta }^3$: this gives a local maximum at $\beta =-2$ with $\mathrm{g}(-2)=-2-\frac{3}{2}-\frac{1}{4}=-\frac{15}{4}$. At $\beta =1$, the sign of ${\mathrm{g}}^{\prime }$ does not change (the double root), so $(1,3)$ is a horizontal point of inflection on the positive branch. The sketch shows the same vertical asymptote and the two additional features compared with $\mathrm{f}$.

Part (ii).

By Vieta's formulae for $x^2+\alpha x+\beta =0$: $u+v=-\alpha$ and $uv=\beta$. Direct substitution gives:$$u+v+\frac{1}{uv}=-\alpha +\frac{1}{\beta },$$$$\frac{1}{u}+\frac{1}{v}+uv=\frac{u+v}{uv}+uv=-\frac{\alpha }{\beta }+\beta .$$Part (iii).

The condition $u+v+\frac{1}{uv}=-1$ gives $-\alpha +\frac{1}{\beta }=-1$, so $\alpha =1+\frac{1}{\beta }$. Substituting into the expression from part (ii):$$\frac{1}{u}+\frac{1}{v}+uv=-\frac{\alpha }{\beta }+\beta =-\frac{1}{\beta }\left(1+\frac{1}{\beta }\right)+\beta =\beta -\frac{1}{\beta }-\frac{1}{{\beta }^2}=\mathrm{f}(\beta ).$$The task reduces to showing $\mathrm{f}(\beta )⩽-1$ whenever $u,v$ are real.

For $u,v$ real the discriminant must be non-negative: ${\alpha }^2-4\beta ⩾0$. Substituting $\alpha =1+1/\beta$:$${\left(1+\frac{1}{\beta }\right)}^2⩾4\beta .$$For $\beta <0$ the right-hand side is negative while the left is non-negative, so the inequality holds automatically for all $\beta <0$. For $\beta >0$, multiply through by ${\beta }^2>0$:$${\beta }^2+2\beta +1⩾4{\beta }^3⟹4{\beta }^3-{\beta }^2-2\beta -1⩽0.$$Factoring: $4{\beta }^3-{\beta }^2-2\beta -1=(\beta -1)(4{\beta }^2+3\beta +1)$. The quadratic $4{\beta }^2+3\beta +1$ has discriminant $9-16<0$ and positive leading coefficient, so it is always positive. Therefore the reality condition for $\beta >0$ is $\beta ⩽1$.

The admissible values are thus $\beta <0$ or $0<\beta ⩽1$. In each region, compare $\mathrm{f}(\beta )$ with $-1$:

- For $\beta <0$: the unique stationary point on this branch is the local maximum $(-1,-1)$ from part (i), so $\mathrm{f}(\beta )⩽-1$ throughout.
- For $0<\beta ⩽1$: ${\mathrm{f}}^{\prime }(\beta )=({\beta }^3+\beta +2)/{\beta }^3>0$ (all terms positive for $\beta >0$), so $\mathrm{f}$ is strictly increasing on this interval with $\mathrm{f}(1)=1-1-1=-1$. Thus $\mathrm{f}(\beta )⩽-1$ for $0<\beta ⩽1$.

In both cases $\frac{1}{u}+\frac{1}{v}+uv=\mathrm{f}(\beta )⩽-1$, as required.

Part (iv).

Now the condition $u+v+\frac{1}{uv}=3$ gives $\alpha =\frac{1}{\beta }-3$, and the same substitution yields:$$\frac{1}{u}+\frac{1}{v}+uv=\beta +\frac{3}{\beta }-\frac{1}{{\beta }^2}=\mathrm{g}(\beta ).$$The problem is now to maximise $\mathrm{g}(\beta )$ subject to the reality condition ${\alpha }^2⩾4\beta$, i.e., $(1/\beta -3{)}^2⩾4\beta$.

For $\beta <0$ the right-hand side is again negative, so the condition holds for all $\beta <0$. For $\beta >0$, multiply by ${\beta }^2$:$$1-6\beta +9{\beta }^2⩾4{\beta }^3⟹4{\beta }^3-9{\beta }^2+6\beta -1⩽0.$$Factoring: $4{\beta }^3-9{\beta }^2+6\beta -1=(\beta -1{)}^2(4\beta -1)$. This is $⩽0$ precisely when $4\beta -1⩽0$ (using $(\beta -1{)}^2⩾0$) or $\beta =1$. So the admissible positive values are $0<\beta ⩽\frac{1}{4}$ together with $\beta =1$.

Evaluate $\mathrm{g}$ on each admissible region:
- $\beta <0$: $\mathrm{g}$ has a local maximum at $\beta =-2$ with $\mathrm{g}(-2)=-\frac{15}{4}$.
- $0<\beta ⩽\frac{1}{4}$: ${\mathrm{g}}^{\prime }(\beta )=(\beta -1{)}^2(\beta +2)/{\beta }^3>0$ throughout, so $\mathrm{g}$ is increasing and attains its maximum at $\beta =\frac{1}{4}$: $$\begin{gathered} \mathrm{g} \\ !\left(\frac{1}{4}\right)=\frac{1}{4}+12-16=-\frac{15}{4} \end{gathered}$$.
- $\beta =1$: $\mathrm{g}(1)=1+3-1=3$.

By reference to the sketch of $\mathrm{g}$ from part (i), the point $(1,3)$ is a horizontal inflection sitting above all other values on the admissible set. The greatest value of $\frac{1}{u}+\frac{1}{v}+uv$ is therefore $\boxed{3}$.