Let x, y be positive reals such that $1 \over x$ + $1 \over y$ $\le 1$... 
Let $x$ and $y$ be positive reals such that $1 \over x$ + $1 \over y$ $\le 1$ and let $a, b ,c$ be the lengths of the sides of a triangle. Show that $a^2x+b^2y\gt c^2.$

I have no idea how to solve this problem. I imagine that given the sides of the triangle the side inequality might be useful, and it`s easy to see that $x,y\gt 1$, but I have not been able to proceed from there. Any suggestions/solutions?
 A: It's wrong. Try $a=b=c=0.001$ and $x=y=2$.
By the way, the following statement is true.
Let $x$ and $y$ be positive numbers such that $\frac{1}{x}+\frac{1}{y}\leq1$ 
and let $a$, $b$ and $c$ be sides-lengths of a triangle.  Prove that:
$$a^2x+b^2y>c^2.$$
Indeed, since $1\leq\frac{xy}{x+y}$, it's enough to prove that
$$a^2x+b^2y>\frac{c^2xy}{x+y}$$ or
$$a^2x^2+(a^2+b^2-c^2)xy+b^2y^2>0,$$
for which it's enough to prove that
$$(a^2+b^2-c^2)^2-4a^2b^2<0,$$ which is
$$\sum_{cyc}(2a^2b^2-a^4)>0,$$ which says that we need to prove that the area of the triangle is positive, 
which is true of course.
Done!
A: I don't understand
the final conclusion in
Michael Rozenberg's answer.
Here is my modification.
From
$(a^2+b^2-c^2)^2-4a^2b^2
\lt 0
$
we get
$(a^2+b^2-c^2-2ab)(a^2+b^2-c^2+2ab)
\lt 0
$
or
$((a-b)^2-c^2)((a+b)^2-c^2)
\lt 0
$
or
$(a-b)^2
\lt c^2
$
(since
$a+b \gt c$).
If $a \le b$,
this is
$a-b < c$
or
$a < b+c$
which is true.
Similarly,
if $a > b$
this is
$b-a < c$
or
$b < a+c$
which is also true.
