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?
On
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.
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!