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\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}

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    $\begingroup$ $a + b \leq |a| + |b| \leq 4$ $\endgroup$
    – Math Lover
    Aug 23 '21 at 1:45
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    $\begingroup$ Alt. hint: by AM-GM $\,ab \le \dfrac{a^2+b^2}{2}\,$, then $\,(a+b)^2 = a^2+b^2+2ab \le 2(a^2+b^2) \le 8\,$. $\endgroup$
    – dxiv
    Aug 23 '21 at 1:56
  • $\begingroup$ You already nearly solved it before the $(a-b)^2$ part. $\endgroup$
    – Saeed
    Aug 24 '21 at 2:24
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If $a + b > 4$, then $a > 2$ or $b > 2$. If, for instance, $a > 2$, then $a^2 +b^2 \geq a^2 > 4$.

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Applying the RMS-AM inequality :

$$ 𝑎^2 \ + \ 𝑏^2 \ ≤ \ 4 \ \ \Rightarrow \ \ \frac{𝑎+𝑏}{2} \ ≤ \ \sqrt{\frac{𝑎^2 + 𝑏^2}{2}} \ ≤ \ \sqrt2 \ \ \Rightarrow \ \ 𝑎+𝑏 \ ≤ \ 2 · \sqrt2 \ < \ 4 \ \ . $$

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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 \ \ . $

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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 $

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Alternatively, $(a+b)^2 = a^2+b^2+2ab \le a^2+b^2+(a^2+b^2) = 2(a^2+b^2) = 2\cdot 4 = 8\implies a+b \le \sqrt{8} < 4$

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By rearrangement inequality we have

$$2ab\le a^2+b^2\le 4$$

and

$$(a+b)^2=a^2+b^2+2ab\le 8\implies |a+b|\le 2\sqrt 2$$

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