# Elementary proofs of inequalities

I was just introduced into elementary proofs of inequalities, my text's explanation however feels incomplete. I did further research on the subject, my question is thus:

Prove: If $0 < a < b$, if $c < d$, and $c > 0$, then:

$$ac < bd$$

I understand that you may add or multiply an inequality by a number; however i cannot seem to determine what to use to show what is desired. Another approach i attempted is the fact that if $a < b$ then there is some k s.t. $a + k = b$. And similar $m$ for $c < d$. However plugging this in for $ac < bd$ proved no avail.

What are some ways i could solve this? I am sure there is more than one way.

• If $0 < a < b$, if $c < d$ and $c>0$ Let $\mathcal{P}^{+}$ be the set of positive numbers. Since $(b-a), c \in \mathcal{P}^{+}$, then $c(b-a)\in P^{+}$, i.e., $ac<bc$. Also $(d-c),b\in \mathcal{P}^{+}$ so $b(d-c)\in \mathcal{P}^{+}$,i.e., $bc<bd$. Then we have $ac<bc$ and $bc<bd$ and hence (by transitivity) $ac<bd$ as desired. Mar 28, 2014 at 2:17

Given: $$$$\tag{P1}0 < a < b$$$$$$$$\tag{P2} 0 < c < d$$$$$$c\times (P1) \implies $$\tag{3} ac < bc$$$$$$b\times (P2)\implies $$\tag{4}bc < bd$$$$$$(3)\land(4)\implies$$\tag{C} ab < bc < bd$$$$$$\therefore ac<bd$$

Using

• if $0 \lt y$ then $x \lt x+y$
• if $x \lt y$ then $0 \lt y-x$
• if $0 \lt x$ and $0 \lt y$ then $0 \lt xy$

you have

$$ac \lt ac+ (b-a)c = bc \lt bc+b(d-c)=bd$$

• You can multiple/add sides of an inequality with a different number? I got the impression such an operation must be the same across the entire inequality Mar 28, 2014 at 1:52
• This is four inequalities and equalities on one line. If you prefer: $$ac \lt ac+ (b-a)c$$ $$ac+ (b-a)c = bc$$ $$bc \lt bc+b(d-c)$$ $$bc+b(d-c)=bd$$ so $ac \lt bd$ Mar 28, 2014 at 7:13

Start with $a<b$. Multiply both sides by the positive $c$ and we get $ac<bc$, and since $c<d$ and $c$ and $d$ are both positive ($c$ is positive, and $d$ is larger than $c$, so we can tell that $d$ is positive),we can tell that $ac<bc<bd$, an we can simplify that as $ac<bd$.

There was nothing wrong with your approach as well.

\begin{aligned}a < b & \implies a + k = b &\text{ for some } k > 0 \\ c < d & \implies c + m = d & \text{ for some } m > 0\end{aligned}

The desired inequality then becomes prove that:

\begin{aligned}& ac & < & bd \\ \text{i.e } & ac & < & (a+k)(c+m) \\ \text{i.e } & ac & < & ac + am + kc + km \\ \text{i.e } & 0 & < & am + kc + km\end{aligned}

which is true, since each term on the RHS is positive ($a, c, k, m > 0$).

Replacing $x, y$ with $x, x + \alpha$ where $\alpha = y - x \ge 0$ is often a useful strategy in proving inequalities. See this for example.