Ta có: $a\le 1 \Rightarrow \sqrt{a+1}\le\sqrt2$Tương tự, suy ra:$ \sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\leq3\sqrt2< 3,5 $
Áp dụng BĐT
Cauchy ta có:$
2.\sqrt{a
+1}.\frac{2}{\sqrt 3}\le
a+ 1
+ \
frac{4}{3}$$2.\sqrt
{b+1}.\fra
c{2}{\sqr
t 3}\le b+ 1+ \
frac{4}{3}$$2.\sqrt{
c+1}
.\
frac{2}{\sqrt
3}\le c+ 1+ \frac{4}{3}$
Dẫn t
ới:$\frac{4}{\sqrt 3}\left ( \s
qrt{a+1}+\sqrt{b+1}+\sqrt{c+1} \r
ight ) \le a
+b+c+3+4=8$
$\Rightarrow \sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}
\le
2\sqrt
3<3,5$