How to think about proofs of inequalities (precalculus)? I know that
$a\leq \lvert a \rvert$ and $\lvert a \rvert =\sqrt{a^2}$, so I can write
$$
a\leq \sqrt{a^2}
$$
But say something like
$$
a\leq \sqrt{a^2+b^2}
$$
Intuitively it's true. But can we really prove this or is it just intuition? I think: "we just added $b^2$ on the right side but nothing on the left side. Therefore the right side must be larger (of equal)". Is this the right way to think about it?
 A: If you do not know that $x\mapsto \sqrt x$ is increasing, note that $\sqrt x\ge 0$ whenever it is defined. Therefore, $0\le x< y$ implies
$$0<y-x=\sqrt y^2-\sqrt x^2=\underbrace{(\sqrt y+\sqrt x)}_{\ge 0}(\sqrt y-\sqrt x), $$
whence
$$ \sqrt y-\sqrt x>0.$$
Now from $b^2\ge 0$, we see $a^2+b^2\ge a^2$ and so finally $\sqrt{a^2+b^2}\ge \sqrt {a^2}$.
A: $$a \leq \sqrt{a^2} \leq \sqrt{a^2+b^2}$$
since square root is an increasing function and we know that $b^2$ is nonnegative.
A: Suppose you have some inequality of the kind $a\le b$, then if you apply an increasing function to both sides the inequality holds, that is, suppose that $f$ is an increasing function, then $f(a)\le f(b)$ if and only if $a\le b$.
If $g$ would be a decreasing function then $g(a)\ge g(b)$ if and only if $a\le b$, you only need to check the definition of increasing and decreasing functions to see that these inequalities holds by definition.
In our case the function $f(x):=x^2$ is increasing for non-negative reals, thus if $a\ge 0$ and $\sqrt{a^2+b^2}\ge 0$ then we have that
$$a\le\sqrt{a^2+b^2}\iff f(a)=a^2\le f(\sqrt{a^2+b^2})=a^2+b^2\tag1$$
where the symbol $\iff$ must be read as "if and only if", and it means that if one thing in one of the sides of the $\iff$ is true (or false) the other is also true (or also false).
Then from $(1)$ we find that $a^2\le a^2+b^2\iff 0\le b^2$ what is true, thus $a\le\sqrt{a^2+b^2}$ is also true. Moreover, if $a<0$ then $-a>0$ and you can show that $-a\le\sqrt{a^2+b^2}$, and because $a<-a$ then your inequality also holds when $a<0$.
