Prove modular inequalities $ab + ac\le a(b+ac)$ and $(a+b)(a+c)\ge a+b(a+c)$

Question

How to prove

$$(a\cdot b)+(a\cdot c)\le a\cdot\big(b+(a\cdot c)\big)$$

and

$$(a+b)\cdot(a+c)\ge a+\big(b\cdot(a+c)\big)\;?$$

I have tried this. Using distributive property, I think we can get

$$a+(b\cdot c) \le (a+b)\cdot(a+c)$$

and

$$a\cdot(b+c) \ge (a\cdot b) + (a\cdot c)\;.$$

Now what should I do?

Answer 1

I guess you have the precondition $a,b,c\geq 1$. Then $$ab + ac = a(b+c) = a(b+1\cdot c)\overset{a\geq 1}{\leq} a(b + ac)$$ and $$(a+b)(a+c) = a\underbrace{(a+c)}_{\geq 1} + b(a+c) \geq a + b(a+c).$$

Answer 2

I am assuming that you are interested in lattice operations on positive integers, so $*$ stands for the greatest common divisor and $+$ for the least common multiple. By uniqueness of factorization it suffices to do this one prime factor at a time. So let $p$ be any prime, and denote by $\nu_p(n)$ the highest power of $p$ that divides the positive integer $n$. So $\nu_p(m*n)=\min\{\nu_p(m),\nu_p(n)\}$ and $\nu_p(m+n)=\max\{\nu_p(m),\nu_p(n)\}$.

Let's look at your first inequality. On the left hand side we have $$ \nu_p((a*b)+(a*c))=\max\{\min\{\nu_p(a),\nu_p(b)\},\min\{\nu_p(a),\nu_p(c)\}\} $$ and on the r.h.s. we have $$ \nu_p(a*(b+(a*c)))=\min\{\nu_p(a),\max\{\nu_p(b),\min\{\nu_p(a),\nu_p(c)\}\}\}. $$ If here $\nu_p(a)$ is $\ge$ than the other two, then both sides simplify to $\max\{\nu_p(b),\nu_p(c)\}$. If $\nu_p(a)$ is $\le$ than the other two, then both sides simplify to $\nu_p(a)$. If $\nu_p(b)\le\nu_p(a)\le\nu_p(c)$, then again both sides simplify to $\nu_p(a)$. Also if $\nu_p(c)\le\nu_p(a)\le\nu_p(b)$ both sides still evaluate to $\nu_p(a)$.

Do the other inequality in the same way. This lattice of integers is modular, so you always get equality.