Generalization of the inequality $ab\leq \frac{1}{2}(a^2+b^2)$

Question

Let $a, b, c, d, e$ are positive real numbers and let $f$ be a negative real number. Do we have $$ \vert abc+def \vert \leq \frac{1}{2}\vert(a^2 +b^2)c+ (d^2+e^2)f \vert ~?$$

I tried to produce a counter example but...

Answer 1

No. You can always choose $f$ so that the right side is zero. By choosing $d$ and $e$ small, you can ensure that the left side is non-zero.

Answer 2

Try $a=b=d=c=1$, $e=2$, $f=-0.4$ and we get

$$0.2\leq 0$$

Answer 3

For a slightly more interesting example, consider $$ a=63,b=16, c=1, \\ d=33, e=56, f=-1 $$ Using the fact that $65$ is the hypotenuse of two distinct integer-sided right triangles.

On the other hand, if we required that all variables are $>0$, I think the inequality would hold.