Find the minimum of $\space\frac{1}{x}+\frac{1}{y}+c\cdot xy\space$ subject to $\space x+y-c=0$ Let $f(x,y):\mathbb{D}\rightarrow\mathbb{R}$ be the function:
$$f(x,y)=\frac{1}{x}+\frac{1}{y}+c\cdot xy\space\space|\space\space c\in(0,\sqrt[4]8)\text{ $\space$constant}$$
$$\mathbb{D}=\{(x,y)\space|\space x>0,\space y>0\}$$
Find the point $P$ where $f$ gets its minimum value, subject to the equality:
$$x+y-c=0$$
I tried:
1) Lagrange multipliers. Unfortunately, they don't seem to help since I get equations I cannot solve (5th degree).  
2) Substituting $y=c-x$ to $f$ in order to solve $\frac{d}{dx}f(x,c-x)=0$. That wasn't helpful either, from the same reasons.
Final Solution: The final solution should be $(\frac{c}{2},\frac{c}{2})$; However I could not figure out how to find it by myself.
Thanks!
 A: We have $$f(x,y)=\frac{c}{xy}+cxy,$$ which is decreasing function of $xy$ on $(0,1)$.
Thus, we need to found a maximal value of $xy$, for which the equation
$$t^2-ct+xy=0$$ has positive roots for $0<c<\sqrt[4]{8}.$
Thus, $c^2-4xy\geq0,$ which says $$xy\leq\frac{1}{4}c^2<\frac{1}{\sqrt2}<1.$$
We see that the maximal value of $xy$ it's $\frac{c^2}{4}$ and occurs for $x=y=\frac{c}{2},$ which says  $$\min f(x,y)=\frac{c}{\frac{c^2}{4}}+\frac{c^3}{4}=\frac{4}{c}+\frac{c^3}{4}.$$
A: $\begin{array}\\
f(x,y)
&=\dfrac{1}{x}+\dfrac{1}{y}+c xy\\
&=\dfrac{c}{x(c-x)}+c x(c-x)\\
&=\dfrac{c+cx^2(c-x)^2}{x(c-x)}\\
&=c\dfrac{1+x^2(c-x)^2}{x(c-x)}\\
f_x(x,y)
&= \dfrac{c^3 x^2 - 4 c^2 x^3 + 5 c x^4 - c - 2 x^5 + 2 x}{x^2 (c - x)^2}
\quad\text{according to Wolfy}\\
&= \dfrac{(c - 2 x) (c x - x^2 - 1) (c x - x^2 + 1)}{x^2 (c - x)^2}
\quad\text{also according to Wolfy}\\
\end{array}
$
so
$f$ is extreme at
$x=\frac{c}{2}$
or
$x^2-cx =\pm 1$
or
$x^2-cx+c^2/4 =c^2/4\pm 1$
or
$(x-c/2)^2 
=c^2/4\pm 1$.
If
$(x-c/2)^2
=c^2/4 -1
$
then,
since
$0 \le c \le \sqrt[4]{8}
$,
$0 \le c^2 \le \sqrt{8}
=2\sqrt{2}
$
so
$c^2/4 
\le \dfrac{\sqrt{2}}{2}
\lt 1$
so this can not be.
If
$(x-c/2)^2
=c^2/4 +1
$
then
$x
=\dfrac{c}{2}\pm\sqrt{\dfrac{c^2}{4} +1}
$.
Since $x > 0$
we must have
$x
=\dfrac{c}{2}+\sqrt{\dfrac{c^2}{4} +1}
$.
If
$x = \dfrac{c}{2}$,
$x = y$
so
$x(c-x)
=\dfrac{c^2}{4}
$
so
$\begin{array}\\
f(x, y)
&=c\dfrac{1+x^2(c-x)^2}{x(c-x)}\\
&=c\dfrac{1+c^4/16}{c^2/4}\\
&=\dfrac{4(1+c^4/16)}{c}\\
&=\dfrac{4}{c}+\dfrac{c^3}{4}\\
\end{array}
$
If
$x 
=\dfrac{c}{2}+\sqrt{\dfrac{c^2}{4} +1}
$,
$y
=c-x
=\dfrac{c}{2}-\sqrt{\dfrac{c^2}{4} +1}
\lt 0
$
which is not allowed.
Therefore 
the extreme value is at
$x=y=\dfrac{c}{2}$
where the value is
$\dfrac{4}{c}+\dfrac{c^3}{4}
$.
