\(VT=a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1}\)
\(=a\sqrt{\left(b+1\right)\left(b^2-b+1\right)}+b\sqrt{\left(c+1\right)\left(c^2-c+1\right)}+c\sqrt{\left(a+1\right)\left(a^2-a+1\right)}\)
\(\le\frac{a\left(b^2+2\right)}{2}+\frac{b\left(c^2+2\right)}{2}+\frac{c\left(a^2+2\right)}{2}\) ( BĐT cô si )
\(=\frac{1}{2}\left(2a+2b+2c+ab^2+bc^2+ca^2\right)\)
\(=3+\frac{ab^2+b^2c+c^2a}{2}\)
Gỉa sử b là số ở giữa .
\(\Rightarrow\left(b-a\right)\left(b-c\right)\le0\)
\(\Leftrightarrow b^2+ca\le bc+ab\)
\(\Leftrightarrow ab^2+ca^2\le abc+a^2b\)
\(\Leftrightarrow ab^2+bc^2+ca^2\le b\left(a+c\right)^2=\frac{\left(b+b\right)\left(a+c\right)\left(a+c\right)}{2}\le\frac{\left(2a+2b+2c\right)^3}{54}=4\)
\(\Rightarrow VT\le3+\frac{4}{2}=5\left(đpcm\right)\)
Dấu \("="\)xảy ra khi \(b=1;c=2;a=0\) và hoán vị