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$4ax^2 + 6bx^2 - 2ax^3 - 2ax^2y - 2by^3 - 4bxy^2 \leq 0$ 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!

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  • $\begingroup$ Getting it always equal to zero is easy. Just pick $a$ and $b$ in such a way that everything on the left hand side gets destroyed. Strictly less than zero for all $x, y$ is impossible to achieve (at least when $a$ and $b$ are not allowed to depend on $x$ and $y$). This requires some proof but I think you can find it. $\endgroup$
    – Vincent
    Commented Feb 14, 2018 at 9:00
  • $\begingroup$ Pick $a=b=0$, and you are done. $\endgroup$ Commented Feb 14, 2018 at 9:01
  • $\begingroup$ Ok, here is a hint for the modified question. First solve it for the special case that $x = y = 1$. Then see how the resulting expression changes if $x$ and $y$ vary but the values of $a$ and $b$ stay the same. $\endgroup$
    – Vincent
    Commented Feb 14, 2018 at 10:43

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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 :-).

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  • $\begingroup$ I just changed the condition where x and y always > 0. Does that change your answer? Thanks! $\endgroup$
    – Phu Nguyen
    Commented Feb 14, 2018 at 10:15

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