Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1^2}{\sqrt{a}}+\frac{2^2}{\sqrt{b}}+\frac{3^2}{\sqrt{c}}\ge\frac{\left(1+2+3\right)^2}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\frac{36}{6}=6\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\frac{1}{\sqrt{a}}=\frac{2}{\sqrt{b}}=\frac{3}{\sqrt{c}}\\\sqrt{a}+\sqrt{b}+\sqrt{c}=6\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{a}=1\\\sqrt{b}=2\\\sqrt{c}=3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=4\\c=9\end{matrix}\right.\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\Leftrightarrow ab+bc+ca=abc\)

\(\frac{a^2}{a+bc}=\frac{a^3}{a^2+abc}=\frac{a^3}{a^2+ab+bc+ca}=\frac{a^3}{\left(a+b\right)\left(a+c\right)}\)

Cô si: 

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+b}{8}\ge3\sqrt[3]{\frac{a^3}{\left(a+b\right)\left(b+c\right)}.\frac{\left(a+b\right)}{8}.\frac{\left(b+c\right)}{8}}=\frac{3a}{4}\)

Tương tự với 2 cục còn lại, công theo vế:

\(VT+\frac{a+b+c}{2}\ge\frac{3}{4}\left(a+b+c\right)\)

\(\Rightarrow VT\ge\frac{a+b+c}{4}\text{ }\left(dpcm\right)\)

\(a+b+c=abc\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=1\)

Đặt \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(VT=\frac{x^2yz}{1+yz}+\frac{xy^2z}{1+zx}+\frac{xyz^2}{1+xy}=\frac{x^2yz}{xy+yz+yz+zx}+\frac{xy^2z}{xy+zx+yz+zx}+\frac{xyz^2}{xy+yz+xy+zx}\)

\(VT\le\frac{1}{4}\left(\frac{x^2yz}{xy+yz}+\frac{x^2yz}{yz+zx}+\frac{xy^2z}{xy+zx}+\frac{xy^2z}{yz+zx}+\frac{xyz^2}{xy+yz}+\frac{xyz^2}{xy+zx}\right)\)

\(VT\le\frac{1}{4}\left(\frac{x^2y}{x+y}+\frac{xy^2}{x+y}+\frac{y^2z}{y+z}+\frac{yz^2}{y+z}+\frac{x^2z}{x+z}+\frac{xz^2}{x+z}\right)\)

\(VT\le\frac{1}{4}\left(xy+yz+zx\right)=\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=\sqrt{3}\)