Show if $a^2+b^2 \le 2$ then $a+b \le 2$ If $a^2+b^2 \le 2$ then show that $a+b \le2$
I tried to transform the first inequality to $(a+b)^2\le 2+2ab$ then $\frac{a+b}{2} \le \sqrt{1+ab}$ and I thought about applying $AM-GM$ here but without result
 A: $(a+b)^2+(a-b)^2=2(a^2+b^2)\leq 4$, so $|a+b|\leq 2$
A: Hint Use Cauchy-Schwarz.
Second solution
$$(a+b)^2=a^2+b^2+2ab \leq 2+2ab$$
You got that far, you are almost there:
By AM-GM
$$\sqrt{a^2b^2} \leq \frac{a^2+b^2}{2}$$
which implies $$2ab \leq a^2+b^2 \leq 2$$
A: Let $a=\sqrt{2}\cos\theta$, $y=\sqrt{2}\sin\theta$. Then $a^2+b^2=2$, and $a+b=\sqrt{2}(\cos\theta+\sin\theta)$, which is a maximum at $\theta=\frac{\pi}{4}$, at which case $a+b=2$. So $a+b\le 2$.
A: Yet another method, inspired by looking at the problem geometrically (try drawing the region $a^2+b^2\leq2$ and the line $a+b=2$): let $s=a+b$, $t=a-b$.  Then $a=\frac12(s+t)$ and $b=\frac12(s-t)$, so $a^2+b^2=\frac14\bigl((s^2+2st+t^2)+(s^2-2st+t^2)\bigr) = \frac12(s^2+t^2)$ and $a+b=s$, so the problem becomes: 

if $s^2+t^2\leq 4$ then show that $s\leq 2$.

But this is manifestly true; $t^2\geq 0$, so if $s^2+t^2\leq 4$ then $s^2\leq 4$ and so $s\leq 2$.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
\begin{align}
a + b&=\pars{a,b}\cdot\pars{1,1}\leq\verts{\pars{a,b}}\verts{\pars{1,1}}
=\root{a^{2} + b^{2}}\root{1^{2} + 1^{2}}\leq\root{2}\root{2} = 2
\end{align}

$$
\imp\quad \color{#66f}{\large a + b \leq 2}
$$

A: Suffices to show if $a^2+b^2 = 2$ then $a+b \leq 2$. From the constraint consider
$$
f(a) = a + \sqrt{2-a^2}
$$
and we need to prove $f(a) \leq 2$ over $[0,\sqrt{2}]$.
$$
f'(a) = 1 - \frac{a}{\sqrt{2-a^2}} \Leftrightarrow a = 1
$$
which is a maximum by 1st derivative test.
Since $f(0), f(1), f(\sqrt{2})$ are $\sqrt{2}, 2, \sqrt{2}$ we have $f(a) \leq 2$ as desired.
A: First note that if $a=0$ or $b=0$, then the question is easy. So assume that both are non-zero. Consider the value $ab$. We can show that we must have $|ab|\le1$. 
Assume by contrary that we have:  $|ab| > 1$. This means $$\frac{1}{|b|}<|a|$$
Expanding: $$0\le(b^2-1)^2$$
we get:$$2\le \frac{1}{b^2}+b^2,$$
Thus, using the original ineqaulity we see:
$$
2\le \frac{1}{b^2}+b^2\lt a^2+b^2\le2
$$
which is a contradiction. Hence,$|ab|\le1$, and we can continue your original inequality, by writing:
$$
a+b\le \sqrt{2+2ab}\le \sqrt{2+2}=2
$$
