$a,b \in \mathbb{R}$
My attempt: $$a^{2}+b^{2}=1 \iff a^{2}+2ab+b^{2}=1+2ab$$ $$(a+b)^2=1+2ab \iff \mid a+b \mid =\sqrt{1+2ab} \iff a+b = \pm\sqrt{1+2ab}$$
But notice that $\sqrt{1+2ab} \in \mathbb{R} \iff 1+2ab \geq 0 \iff 2ab \geq -1$ .
let's take the case where :$2ab=-1$. $$a^{2}+2ab+b^{2}=0 \iff a+b=0 \iff a=-b.$$
When I plug this result into the original equation, I get: $$b=\frac{1}{\sqrt{2}}$$ And now, I don't know where to go.
Here is an alternative route:
By the AM–GM inequality, we have $$ {\frac {a+b}2}\geq {\sqrt {ab}} $$ and so $a^2+b^2+2ab \ge 4ab$, which implies $2ab \le 1$. Then $$ (a+b)^2 = a^2+b^2+2ab\le 2 $$ gives $$ |a+b| \le \sqrt 2 $$ as required.
Note that the inequality you wanted to prove is equivalent to $(a+b)^2\leq 2$, so once you've obtained that $(a+b)^2=1+2ab$, the next step should be trying to prove that $1+2ab \leq 2$. This can be proven using the AM-GM inequality, which says that $2ab \leq a^2+b^2$, where the equality is taken when $a=b$. Thus $1+2ab \leq 1+a^2+b^2 = 2$.
In your attempt, you've actually reversed the logic a little bit by first declaring that $\sqrt{1+2ab}$ is real, then trying to prove the condition for its real-ness. This doesn't make sense. You can prove that $2ab\geq -1$ by rearranging $(a+b)^2\geq 0$ and using the fact that $a^2+b^2 = 1$. The equality here is taken when $a = -b$, which leads to the values $b = \pm \frac{1}{\sqrt{2}}$ that you solved for. In any case, this doesn't help you prove the original inequality, because it provides a lower-bound, not an upper bound to $(a+b)^2$.
Actually you don't need to focus on the case $2ab=-1$. Since we know that $|a+b| = \pm \sqrt{1+2ab}$, we just need to prove that $2ab \leqslant 1$.
To do this, observe that
$$ 1-2ab = a^2+b^2-2ab = (a-b)^2 \geqslant 0 $$
Hint (1): Why did you choose the case $2ab = -1$ ? Well, as you showed, this is the smallest possible value for $2ab$ . What can you say about $|a+b|$ when $2ab > -1$ ?
Hint (2) for a different method: Consider a circle of radius 1 around the center of coordinates. The equation of this circle is $x^2 + y^2 = 1$ . Now consider the set of parallel lines $x+y = c$ where $c$ can be any number. Some of these lines touch the circle. Among those lines, which one has the highest value of $c$ ?
Using the triangle inequality and the equivalence between the $\ell^1$ and the $\ell^2$ norm, we have $$|a+b| \leq |a|+|b| = \|(a,b)\|_1 \leq \sqrt{2}\, \|(a,b)\|_2 = \sqrt{2}\,\sqrt{a^2+b^2} = \sqrt{2}.$$
As $a^2 + b^2 = 1$ you can find $\theta \in [0,2\pi]$ such that $a = \cos(\theta)$ and $b = \sin(\theta)$. Now you can study the variations of the function $f : [0,2\pi] \to \mathbb{R}$ defined by $f(\theta) = \cos(\theta) + \sin(\theta)$ and you will see that for all $\theta$, $|f(\theta)| \leq \sqrt{2}$.
Since $a^2+b^2\ge2ab$
$\bigg(a^2+b^2-2ab=\left(a+b\right)^2\ge0\bigg)$
$2ab\le1$.
So, $a^2+b^2+2ab\le2$
$(a+b)^2\le2$.
When you square root both sides, there will be two equations
$a+b\le\sqrt2$
$a+b\ge-\sqrt2$
Combining them gives
$-\sqrt2\le a+b\le\sqrt2$
I'm new to MathJax so please forgive me for weird signs.
$-\sqrt{2}\le a+b\ge \sqrt{2}$. You also need to put a space between a command (that starts with a backslash) and numbers or letters that follow it, for example, instead of \le2, use \le 2. If the command takes an argument, for example when using \sqrt you need to specify the stuff that goes under the radical as an argument, you should enclose it in a pair of curly braces, like this \sqrt{2}.
– Elliot Yu
Feb 02 '21 at 22:48