Nice inequality

Let $\large x_1=\sqrt{4a+1}$

and $\large x_2=\sqrt{4b+1}$

Then by using $\large RMS\geq AM$ inequality on $\large x_1$ and $\large x_2$ ,

we get $\large \sqrt{\frac{x_1^2+x_2^2}{2}}\geq \frac{x_1+x_2}{2}$

So, $\large \sqrt{\frac{4a+1+4b+1}{2}}\geq \frac{\sqrt{4a+1}+\sqrt{4b+1}}{2}$

$\large \Rightarrow \sqrt{\frac{4(a+b)+2}{2}}\geq \frac{\sqrt{4a+1}+\sqrt{4b+1}}{2}$

Since $\large a+b=1$ ,

$\large \Rightarrow \sqrt{\frac{4+2}{2}}\geq \frac{\sqrt{4a+1}+\sqrt{4b+1}}{2}$

$\large \Rightarrow \sqrt{3}\geq \frac{\sqrt{4a+1}+\sqrt{4b+1}}{2}$

$\large \Rightarrow 2\times\sqrt{3}\geq \sqrt{4a+1}+\sqrt{4b+1}$

So the maximum value of $\large \sqrt{4a+1}+\sqrt{4b+1}$ is equal to $\large 2\sqrt{3}$ $\large \approx \boxed{3.464}$

Relevant Wiki : Cauchy Schwartz Inequality

Let $a_{1} = \sqrt{4a+1}$ $a_{2} = \sqrt{4b+1}$ $b_{1} = 1$ $b_{2} = 1$

Then by Cauchy schwarz inequality,

$(\sqrt{4a+1}^2 + \sqrt{4b+1}^2) \cdot (1^2+1^2) \geq (\sqrt{4a+1} \times 1 + \sqrt{4b+1} \times 1)$

$(4(a+b)+2) \cdot 2 \geq (\sqrt{4a+1} + \sqrt{4b+1})^2$

$(\sqrt{4a+1} + \sqrt{4b+1}) \leq (\sqrt{12})$

$\leq 2\sqrt{3}$

$\approx \boxed{3.464}$

$f(a,b)=\sqrt{4a+1}+\sqrt{4b+1}=\sqrt{4a+1}+\sqrt{5-4a}\\ \therefore~f'=\dfrac{1/2*4}{\sqrt{4a+1}} - \dfrac{1/2*4}{\sqrt{5-4a}}=0~for~extrimum.\\ So~\sqrt{4a+1}=\sqrt{5-4a}.~~\implies~ 4a+1=5-4a.\\ So~a=1/2,~and~b=1/2.\\ So~f(1/2,1/2)=2*\sqrt{4*1/2+1}=2*\sqrt3=3.464.\\ Ifa=0,~b=1,~~f=1+\sqrt5=2.236<3.464,~~~so~3.464~is~f_{max}.\\ or\\ Since~it~ is~symmetrical~function~etrimum~at~a=b.~\therefore~a=b=1/2.\\ So~f(1/2,1/2)=2*\sqrt{4*1/2+1}=2*\sqrt3=3.464.$