Áp dụng BĐT Cô-si ta có $2.\sqrt{a+1}.\frac{2}{\sqrt 3}\le a+ 1+ \frac{4}{3}$$2.\sqrt{b+1}.\frac{2}{\sqrt 3}\le b+ 1+ \frac{4}{3}$$2.\sqrt{c+1}.\frac{2}{\sqrt 3}\le c+ 1+ \frac{4}{3}$Suy ra$\frac{4}{\sqrt 3}\left ( \sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \right ) \le a+b+c+3+4=8$$\Rightarrow \sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \le 2\sqrt 3 <3,5$
Áp dụng BĐT Cô-si ta có $2.\sqrt{a+1}.\frac{2}{\sqrt 3}\le a+ 1+ \frac{4}{3}$$2.\sqrt{b+1}.\frac{2}{\sqrt 3}\le b+ 1+ \frac{4}{3}$$2.\sqrt{c+1}.\frac{2}{\sqrt 3}\le c+ 1+ \frac{4}{3}$Suy ra$\frac{4}{\sqrt 3}\left ( \sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \right ) \le a+b+c+3+12=16$$\Rightarrow \sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \le 4\sqrt 3$Như vậy bđt trên là sai với $a=b=c=1/3$
Áp dụng BĐT Cô-si ta có $2.\sqrt{a+1}.\frac{2}{\sqrt 3}\le a+ 1+ \frac{4}{3}$$2.\sqrt{b+1}.\frac{2}{\sqrt 3}\le b+ 1+ \frac{4}{3}$$2.\sqrt{c+1}.\frac{2}{\sqrt 3}\le c+ 1+ \frac{4}{3}$Suy ra$\frac{4}{\sqrt 3}\left ( \sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \right ) \le a+b+c+3+
4=
8$$\Rightarrow \sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \le
2\sqrt 3
<
;3
,5$