Is this inequality property true?

Question

I'm having some trouble defining weather this inequality is true or not...

Basically, I wanted to know if its true that

if $a \geq b$ and $c \geq d \Rightarrow a + c \geq b + d$

Well, basically I just wanted to know if this implication is true:

$n(4n^2 + 42n + 120) + 50 \geq n(n^2 + 13n + 38) + 36$, $\forall n \in \mathbb{N}$

Because $50 \geq 36 \Rightarrow n(4n^2+42n+120) \geq n(n^2+13n+38) $

Is this implication really true?

Thanks!

Answer 1

By definition $a + c \geq b + d\iff a+c-b-d\ge 0\iff (a-b)+(c-d)\ge 0$, but the last expression is a sum of two non-negative entities, so indeed, $a + c \geq b + d$.

As to your implication, while it holds (truth implies truth is deducible in logic), the reasoning is off. You can't deduce from $a + c \geq b + d\& a\ge b$ that $c\ge d$ (there're obvious counterexamples), you need to use external arguments.

You can prove that$$ n(4n^2 + 42n + 120) + 50 \geq n(n^2 + 13n + 38) + 36$$ by comparing coefficients at different powers of $n$ and taking in account that $n\ge 0$.

Answer 2

Yes, the implication is true. $$a\geq b\implies a+c\geq b+c$$ $$c\geq d\implies b+c \geq b+d$$

By transitivity, $$a+c\geq b+d$$