$4ax^2 + 6bx^2 - 2ax^3 - 2ax^2y - 2by^3 - 4bxy^2 \leq 0$ With $x, y > 0$
Is there a way to pick $a$ and $b$ so that the inequality above always holds? If there are $a$ and $b$ that can make that statement true, please show me how to prove it. I have been trying for a few hours, but I'm getting nowhere. Not including the trivial case where both a and b = 0. Help please! Thank you!
Very quick answer: no, you can't, because if you take $(x,y)=(0,0)$, then you have the expression equal to $0$, and not strictly negative.
Let's tackle the case where equality to $0$ is allowed...
You need to do it step by step.
Let's place ourselves in the case $y=0$.
Then you need to have:
$(4a+6b)x^2-2ax^3<0$
Meaning that you need also $a=0$, to negate the $x^3$ term that can take positive and negative values and diverge quiker than $x^2$.
Then you need $b<0$, for obvious reasons...
BUT, let's take then $x=0$...
You get with the same reasonning that you need to have $b=0$.
Obviously the equation $0=0$ is true, whatever $(x,y)$, but it is not very interesting :-).