\(a+b+c=3\\ \Leftrightarrow a\left(b+c+2\right)=ab+ac+a+b+c+1=\left(a+1\right)\left(b+c+1\right)\)
Tương tự:
\(b\left(c+a+2\right)=\left(b+1\right)\left(a+c+1\right)\\ c\left(a+b+2\right)=\left(c+1\right)\left(a+b+1\right)\)
Áp dụng BĐT cosi:
\(\left\{{}\begin{matrix}\left(a+1\right)\left(b+c+1\right)\le\dfrac{\left(a+1+b+c+1\right)^2}{2}=\dfrac{2^2}{2}=2\\\left(b+1\right)\left(a+c+1\right)\le\dfrac{\left(b+1+a+c+1\right)^2}{2}=\dfrac{2^2}{2}=2\\\left(c+1\right)\left(a+b+1\right)\le\dfrac{\left(c+1+a+b+1\right)^2}{2}=\dfrac{2^2}{2}=2\end{matrix}\right.\)
Cộng vế theo vế 2 BĐT trên:
\(\Leftrightarrow\sqrt{a\left(b+c+2\right)}+\sqrt{b\left(c+a+2\right)}+\sqrt{c\left(a+b+2\right)}\le2+2+2=6\)
Dấu \("="\Leftrightarrow a=b=c=1\)
Áp dụng BĐT Bunhiacopski:
\(VT^2=\left(\sqrt{a\left(b+c+2\right)}+\sqrt{b\left(a+c+2\right)}+\sqrt{c\left(a+b+2\right)}\right)^2\\ \le\left(a+b+c\right)\left(b+c+2+a+c+2+a+b+2\right)\\ =3\cdot\left(2\cdot3+6\right)=36\\ \Leftrightarrow VT\le\sqrt{36}=6\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{b+c+2}}{\sqrt{a}}=\dfrac{\sqrt{a+c+2}}{\sqrt{b}}=\dfrac{\sqrt{a+b+2}}{\sqrt{c}}\\a+b+c=3\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=1\)