# If $a^{2}+b^{2} \leq 4$, prove that $a+b \leq 4$

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

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

$$$$(a-b)^{2}=a^{2}+b^{2}-2 a b \geq 0$$$$$$$$a^{2}+b^{2} \geq 2 a b \text { So at this step I am not certain what to do next. }$$$$

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

If $$a + b > 4$$, then $$a > 2$$ or $$b > 2$$. If, for instance, $$a > 2$$, then $$a^2 +b^2 \geq a^2 > 4$$.

Applying the RMS-AM inequality :

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

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 \ \ .$$

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 $$-\sqrt{8} \le a + b \le \sqrt{8} \lt 4$$

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

By rearrangement inequality we have

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

and

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