Infimum and supremum of $\;\;\frac{a+b}{\sqrt{a^2+b^2}}$

         Following Infimum and supremum we will prove

$$1=\underset{a>0,b>0}{inf} \;\frac{a+b}{\sqrt{a^2+b^2}}<\frac{a+b}{\sqrt{a^2+b^2}}\leqslant \underset{a>0,b>0}{sup }\;\frac{a+b}{\sqrt{a^2+b^2}}=\frac{a+b}{\sqrt{a^2+b^2}}\mid \underset{(a,b)=(\delta,\delta){\mid}\delta>0}{}=\sqrt{2}$$

           For the inequality on the right we have, for $a>0,\;b>0$

$(a-b)^2 \geqslant 0\;\Rightarrow\;2ab \leqslant a^2+b^2\;\Rightarrow \;a^2+b^2+2ab\leqslant 2(a^2+b^2)$

$\;\Rightarrow\; (a+b)^2\leqslant 2\cdot (a^2+b^2)\;\Rightarrow\;a+b \leqslant \sqrt{2}\cdot \sqrt{a^2+b^2}\;\Rightarrow\;\frac{a+b}{\sqrt{a^2+b^2}}\leqslant \sqrt{2};$

the $=$ sign occurs when $a=b$.

            For the inequality on the left we have, for $a>0,b>0$

$0<2ab\;\Rightarrow \;a^2+b^2<a^2+b^2+2ab\;\Rightarrow a^2+b^2<(a+b)^2\;\Rightarrow\;$  

$\Rightarrow\;\sqrt{a^2+b^2}<a+b\;\Rightarrow\;1<\frac{a+b}{\sqrt{a^2+b^2}}.$ The value $1$ of the fraction $\frac{a+b}{\sqrt{a^2+b^2}}$ is never reached, but whatever $\varepsilon >0$ is, the fraction takes for certain $a_{\varepsilon}$ and $b_{\varepsilon}$ the value $1+\varepsilon$.

     Indeed, let $a_{\varepsilon}=1+(1+\varepsilon)\sqrt{1-2\varepsilon-\varepsilon^2}$ and $b=2\varepsilon+\varepsilon^2$. We have 

$a_{\varepsilon}+b_{\varepsilon}=1+2\varepsilon+\varepsilon^2+(1+\varepsilon)\sqrt{1-2\varepsilon-\varepsilon^2})=(1+\varepsilon)(1+\varepsilon+\sqrt{1-2\varepsilon-\varepsilon^2})$

and

$a_{\varepsilon}^2+b_{\varepsilon}^2=1+(1+\varepsilon)^2(1-2\varepsilon-\varepsilon^2)+2(1+\varepsilon)\sqrt{1-2\varepsilon-\varepsilon^2}+(2\varepsilon+\varepsilon^2)^2=$

$\underset{\alpha=2\varepsilon+\varepsilon^2}{=}1+(1+\alpha)(1-\alpha)+2(1+\varepsilon)\sqrt{1-\alpha}+\alpha^2=2+2(1+\varepsilon)(\sqrt{1-\alpha}=$

$=1+\alpha+2(1+\varepsilon)\sqrt{1-\alpha}+1-\alpha \;\;\underset{1+\alpha=(1+\varepsilon)^2}{=}\;(1+\varepsilon)^2+2(1+\varepsilon)\sqrt{1-\alpha}+(1-\alpha)=$

$=(1+\varepsilon+\sqrt{1-\alpha})^2.$

 So, $\sqrt{a_{\varepsilon}^2+b_{\varepsilon}^2}=1+\varepsilon+\sqrt{1-\alpha}=1+\varepsilon+\sqrt{1-2\varepsilon-\varepsilon^2}.$

     We finally get $\frac{a_{\varepsilon}+b_{\varepsilon}}{\sqrt{a_{\varepsilon}^2+b_{\varepsilon}^2}}=\frac{(1+\varepsilon)(1+\varepsilon+\sqrt{1-2\varepsilon-\varepsilon^2}}{1+\varepsilon+\sqrt{1-2\varepsilon-\varepsilon^2}}=1+\varepsilon$.

$\blacksquare \blacksquare$