Olympiad Math Problem: Find the maximum and minimum of P (Heuristic Solution)

Find the minimum and maximum of $P=\dfrac{y−x}{x+8y}$ for all real $x$ and $y$ that satisfy the equation $y^2(6-x^2)-xy-1=0$.

My solution:

Note that the given equality $y^2(6-x^2)-xy-1=0$ has the terms $y^2,\,x^2y^2$ and $xy$ while the target expression, $P$ is a rational function with the terms $x$ and $y$.

That gives us a nice hint of what can be done, i.e. to rewrite $P$ as a function defined only by the variable $xy$:

First, note that we could rewrite the given inequality in three different manners:

$y^2(6-x^2)-xy-1=0$

$\displaystyle \color{yellow}\bbox[5px,purple]{y^2=\dfrac{xy+1}{6-x^2}}$

$\displaystyle \color{yellow}\bbox[5px,green]{6-x^2=\dfrac{xy+1}{y^2}\,\,\implies x^2=6+\dfrac{xy+1}{y^2}}$

$\displaystyle \color{black}\bbox[5px,orange]{6y^2-x^2y^2-xy-1=0\,\,\implies y^2=\dfrac{x^2y^2+xy+1}{6}}$

Now, looking back at the target function $P$ and judging from the three different functions we have gotten as shown above, it's quite clear that we should multiply both top and bottom of $P$ by $y$ such that

$\begin{align*}P&=\dfrac{y−x}{x+8y}\\&=\dfrac{y(y−x)}{y(x+8y)}\\&=\dfrac{y^2−xy}{xy+8y^2}\\&=\dfrac{\dfrac{x^2y^2+xy+1}{6}−xy}{xy+8\left(\dfrac{x^2y^2+xy+1}{6}\right)}\\&=\dfrac{x^2y^2+xy+1-6xy}{6xy+8x^2y^2+8xy+8}\\&=\dfrac{x^2y^2-5xy+1}{8x^2y^2+14xy+8}\\&=\dfrac{x^2y^2-5xy+1}{8(x^2y^2+\dfrac{14xy}{8}+1)}\\&=\dfrac{1}{8}\left(1-\dfrac{27xy}{4(x^2y^2+\dfrac{14xy}{8}+1)}\right)\\&=\dfrac{1}{8}\left(1-\dfrac{27}{4(\dfrac{14}{8}+xy+\dfrac{1}{xy})}\right)\end{align*}$

If we want to look for $P_{\text{minimum}}$, we need $\displaystyle \color{black}P_{\text{minimum}}=\dfrac{1}{8}\left(1-\dfrac{27}{4\left(\color{yellow}\bbox[5px,purple]{\dfrac{14}{8}+xy+\dfrac{1}{xy}}\right)}\right)$ where we need a minimum $\displaystyle \color{yellow}\bbox[5px,purple]{\dfrac{14}{8}+xy+\dfrac{1}{xy}}$.

If $xy\gt 0$, then we see that we have $xy+\dfrac{1}{xy}\ge 2$, equality attains when $xy=1$, therefore we get:

$\dfrac{27}{4(\dfrac{14}{8}+xy+\dfrac{1}{xy})}\ge \dfrac{27}{4(\dfrac{14}{8}+2)}=\dfrac{9}{5}$

So we get $P_{\text{minimum}}=\dfrac{1}{8}\left(1-\dfrac{9}{5}\right)=-\dfrac{1}{10}$.


If $xy\lt 0$, then we need to rewrite $P$ such that it takes the form:

$\begin{align*}P&=\dfrac{1}{8}\left(1-\dfrac{27}{4(\dfrac{14}{8}+xy+\dfrac{1}{xy})}\right)\\&=\dfrac{1}{8}\left(1-\dfrac{27}{4(\dfrac{14}{8}-|xy|-\dfrac{1}{|xy|})}\right)\\&=\dfrac{1}{8}\left(1+\dfrac{27}{4\left(|xy|+\dfrac{1}{|xy|}-\dfrac{14}{8}\right)}\right)\end{align*}$

If we want to look for $P_{\text{maximum}}$, we need $\displaystyle \color{black}P_{\text{maximum}}=\dfrac{1}{8}\left(1+\dfrac{27}{4\left(\color{yellow}\bbox[5px,green]{|xy|+\dfrac{1}{|xy|}-\dfrac{14}{8}}\right)}\right)$ where we need a minimum $\displaystyle \color{yellow}\bbox[5px,green]{|xy|+\dfrac{1}{|xy|}+\dfrac{14}{8}}$.

From here we get $\dfrac{27}{4\left(|xy|+\dfrac{1}{|xy|}-\dfrac{14}{8}\right)}\ge \dfrac{27}{4\left(2-\dfrac{14}{8}\right)}=27$

Thus $P_{\text{maximum}}=\dfrac{1}{8}(1+27)=\dfrac{7}{2}$.

Can you see it now? With a bit of the algebraic manipulation, AM-GM inequality can be such a huge helper to simplify the optimization problem and it's like a short cut to solve a pretty headache problem! :D