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}

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 $
