if $a^{2}+b^{2}=1$ Prove that $-\sqrt{2} \le a+b \le \sqrt{2}$ $a,b \in \mathbb{R}$
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.
 A: 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.
A: 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 $(a+b)^2$.
A: 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
$$ 1-2ab = a^2+b^2-2ab = (a-b)^2 \geqslant 0 $$
A: 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 $c$ ?
A: Using the triangle inequality and the equivalence between the $\ell^1$ and the $\ell^2$ norm, we have
$$|a+b| \leq |a|+|b| = \|(a,b)\|_1 \leq \sqrt{2}\, \|(a,b)\|_2 = \sqrt{2}\,\sqrt{a^2+b^2} = \sqrt{2}.$$
A: 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 $\theta$, $|f(\theta)| \leq \sqrt{2}$.
A: 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.
