Nhân 2 vế của giả thiết với \(abc\) ta có: \(ab+bc+ca=abc\)

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

Áp dụng BĐT AM-GM ta có: 

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3a}{4}\)

Tương tự cho 2 BĐT còn lại ta cũng có:

\(\frac{b^2}{b+ca}+\frac{b+c}{8}+\frac{b+a}{8}\ge\frac{3b}{4};\frac{c^2}{c+ab}+\frac{c+a}{8}+\frac{c+b}{8}\ge\frac{3c}{4}\)

Cộng theo vế 3 BĐT trên ta có: 

\(VT+\frac{4a+4b+4c}{8}\ge\frac{3a+3b+3c}{4}\)

\(\Leftrightarrow VT+\frac{2a+2b+2c}{4}\ge\frac{3a+3b+3c}{4}\Leftrightarrow VT\ge VP\)

Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)

\(\Rightarrow ab+bc+ca=abc\)

\(VT=\frac{a^3}{a^2+abc}+\frac{b^3}{b^2+abc}+\frac{c^3}{c^2+abc}\)

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

\(VT=\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{b^3}{\left(b+c\right)\left(a+b\right)}+\frac{c^3}{\left(a+c\right)\left(b+c\right)}\)

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\hept{\begin{cases}\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\\\frac{b^3}{\left(b+c\right)\left(a+b\right)}+\frac{b+c}{8}+\frac{a+b}{8}\ge3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\\\frac{c^3}{\left(a+c\right)\left(b+c\right)}+\frac{a+c}{8}+\frac{b+c}{8}\ge3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\end{cases}}\)

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

\(\Rightarrow\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{b^3}{\left(b+c\right)\left(a+b\right)}+\frac{c^3}{\left(a+c\right)\left(b+c\right)}\ge\frac{a+b+c}{4}\)

\(\Leftrightarrow\frac{a^2}{a+bc}+\frac{b^2}{b+ca}+\frac{c^2}{c+ab}\ge\frac{a+b+c}{4}\left(đpcm\right)\)

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Ta có:

\(\hept{\begin{cases}a^2+b^2+c^2=14\\a+2b+3c=14\end{cases}\Leftrightarrow\hept{\begin{cases}a^2+b^2+c^2=14\\2a+4b+6c=28\end{cases}}}\)

\(\Rightarrow a^2+b^2+c^2-2a-4b-6c=-14\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-3\right)^2=0\)

\(\Leftrightarrow a=1;c=2;c=3\)

Vậy \(T=abc=6\)

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