If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$ \begin{equation}
\text { If } a^{2}+b^{2} \leq 4 \text { prove that } a+b \leq 4 \text { }
\end{equation} What I have tried:

\begin{equation}
a^{2} \leq a^{2}+b^{2} \leq 4 \text { and } b^{2} \leq a^{2}+b^{2} \leq 4
\end{equation}
\begin{equation}
|a| \leq 2 \text { and }|b| \leq 2
\end{equation}
\begin{equation}\text { So } a+b 
\text { can get the maximum value, then } a \geq 0 \text { and } b \geq 0 \text { }
\end{equation}\begin{equation}
\text {  } 0 \leq a \leq 2 \text { and } 0 \leq b \leq 2 \text {}
\end{equation}

\begin{equation}
(a-b)^{2}=a^{2}+b^{2}-2 a b \geq 0
\end{equation}\begin{equation}
a^{2}+b^{2} \geq 2 a b \text {  So at this step I am not certain what to do next. }
\end{equation}
 A: From Cauchy-Schwartz inequality,
$a^2 + b^2 = (a^2 + b^2) ( \dfrac{1}{2} + \dfrac{1}{2} ) \ge (\dfrac{1}{\sqrt{2}} a + \dfrac{1}{\sqrt{2}} b )^2 = \dfrac{1}{2} (a + b)^2$
Hence $(a + b)^2 \le 2 (a^2 + b^2) = 8 $
Thus $|a + b| \le \sqrt{8} $
The last inequality is equivalent to $ -\sqrt{8} \le a + b \le \sqrt{8} \lt 4 $
A: Alternatively, $(a+b)^2 = a^2+b^2+2ab \le a^2+b^2+(a^2+b^2) = 2(a^2+b^2) = 2\cdot 4 = 8\implies a+b \le \sqrt{8} < 4$
A: By rearrangement inequality we have
$$2ab\le a^2+b^2\le 4$$
and
$$(a+b)^2=a^2+b^2+2ab\le 8\implies |a+b|\le 2\sqrt 2$$
A: If $a + b > 4$, then $a > 2$ or $b > 2$. If, for instance, $a > 2$, then $a^2 +b^2 \geq a^2 > 4$.
A: Applying the RMS-AM inequality  :
$$ ^2 \ + \ ^2 \ ≤ \ 4  \ \ \Rightarrow \ \    \frac{+}{2} \ ≤ \ \sqrt{\frac{^2  + ^2}{2}} \ ≤ \ \sqrt2  \ \ \Rightarrow \ \  +  \ ≤ \ 2 · \sqrt2 \ < \ 4 \ \ . $$

Another argument:  the circle of radius $ \ 2 \ $ centered on the origin and its interior, as described by $ \ x^2 + y^2 \ \le \ 4 \ \ , $ lies entirely "below" the line $ \ x + y = \ 4 \ \ . $
