Let $a,b,c,d$ be positive real numbers such that $a+b+c+d=1$, then prove that $$S=\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1} < 6$$
Source : Inequalities (Page Number 197)
I used Cauchy-Schwarz Inequality on these two sets : $\{\sqrt{4a+1},\sqrt{4b+1},\sqrt{4c+1},\sqrt{4d+1}\}$ and $\{1,1,1,1\}$ to get:
$$S^2 \leq \big(4a+1+4b+1+4c+1+4d+1\big)(4) = \big(4(a+b+c+d) + 4\big)(4)=32$$
$$\implies S \leq \sqrt{32} = 4\sqrt{2} \approx 5.6568$$
So, I am able to prove a somewhat "sharper" inequality than the one given.
Just out of curiosity, I wanted to know if there is a different and preferably more elegant method of tackling this problem.
How to actually prove the original inequality ?
Thanks in advance ! :)