Why do inequalities matter and what are they used for? I am currently studying mathematical course at my college, and I cannot seem to grasp the concept of inequalities.
What troubles me is that, like it's said, "triangle inequality matters because many other theorems are dependent of it". But I have no idea why triangle inequality matters, why bernoullie's inequality or why sin function inequalities matter?
I would be grateful if anyone explained all of this. Practical examples would be much appreciated.
 A: Premise: this is a really general answer and more "intuitive" than practical.
Does it matter if I told you that the amount of money you have in the bank is positive? Sometimes even if we cannot give an exact value to a variable, it is still useful to know that it is larger (or smaller) than some quantity. In real life, we may react differently if something is larger or smaller than a certain value (e.g., if your body temperature is above 41 degrees Celsius, then you DO need to go to the emergency room). Similarly, in mathematics, if some variable is larger than another, then some particular effect could be triggered.
A rather simple example: when you take a square root, you need the number to be non-negative (assuming we're dealing with real numbers). If the quantity under the square root is a complicated expression $f(x)$, involving, say, one variable $x$, then it's hard to figure out if a particular value of $x$ will lead to a meaningful square root. To avoid to evaluate this rather complicated expression every time you're given a new value for $x$, you may manipulate the expression $f(x)\geq 0$, to try to get some bounds directly on $x$. For instance, if you can get to something like $a\leq x \leq b$, with $a$ and $b$ given, then it is easier to verify if a given value for $x$ is "acceptable".
Sometimes in math it is not easy to establish "directly" that a given quantity satisfies a certain inequality. Therefore, we often use "intermediate" inequalities, just like sometimes you use the quadratic formula as "intermediate step" while solving a complex equation.
A: An example of the triangle inequality, and the submultiplicativity of a C$^\star$-norm, being useful is the following theorem:  

Let $\mathcal{A}$ be a C$^\star$-algebra. If $a \in \mathcal{A}$ and $\|a \|<1$, then $1- a$ is invertible with $$(1-a)^{-1} = \sum_{n=0}^\infty a^n.$$

With the triangle inequality we have $$\| \sum_{i=0}^n a^i - \sum_{i=0}^m a^i\| \le \sum_{i=n+1}^m \| a^i \| \le \sum_{1=n+1}^m \| a\|^i \underset{\to}{n,m \to \infty} 0. $$
So in particular the element $\sum_{i=0}^\infty a^i$ exists in $\mathcal{A}$ and we have 
$$ \| 1 - (1-a)\sum_{i=0}^n a^i\| = \| 1 - (1 - a^{n+1} ) \| = \| a^{n+1} \| \le \| a \|^{n+1} \to 0.$$
Thus indeed $(1- a)^{-1}= \sum_{i=0}^\infty a^i $.
A: To give a simple example of when an inequality can be helpful, consider the sequence $$\left\{\frac{\sin(\frac{1}{n})}{n}\right\}_{n = 1}^{\infty}$$
When I first saw this sequence I had no idea how to prove whether it converged or not. The classic way to show it does in fact converge is to realize that $$\left|\sin(\frac{1}{n})\right| \leqslant 1, \text{ for all } n \in \mathbb{N}.$$
This allows you to bound the original sequence as follows,
$$\frac{-1}{n} \leqslant \frac{\sin(\frac{1}{n})}{n} \leqslant \frac{1}{n}, \text{ for all } n \in \mathbb{N}$$
and now it is rather intuitive that the sequence converges to $0$, and in fact follows from the squeeze theorem.
Perhaps this is too simple of an example to be helpful, but I think it highlights the typical use of inequalities in analysis. What we really need in order to prove that this sequence converges to $0$ is a way to "control" the sequence when $n$ starts to get large. Directly controlling this sequence would be difficult as $\sin(\frac{1}{n})$ varies wildly as $n$ becomes large, but instead we can use the inequalities coming from the bound on $\sin$ to control the sequence, and this makes it much easier.
