Giải thích các bước giải:
Ta sử dụng 2 BĐT sau:
$+) {BDT:Cauchy - Schwarz}$: $\dfrac{{{a^2}}}{x} + \dfrac{{{b^2}}}{y} + \dfrac{{{c^2}}}{z} \ge \dfrac{{{{\left( {a + b + c} \right)}^2}}}{{x + y + z}}$
$ + )BDT\left( 1 \right):{\left( {a + b + c} \right)^2} \ge ab + bc + ac,\forall a,b,c$
Ta có:
$\begin{array}{l}
\dfrac{1}{{{a^2} + {b^2} + {c^2}}} + \dfrac{{2009}}{{ab + bc + ac}}\\
= \left( {\dfrac{1}{{{a^2} + {b^2} + {c^2}}} + \dfrac{1}{{ab + bc + ac}} + \dfrac{1}{{ab + bc + ac}}} \right) + \dfrac{{2007}}{{ab + bc + ac}}\\
\ge \dfrac{{{{\left( {1 + 1 + 1} \right)}^2}}}{{{a^2} + {b^2} + {c^2} + ab + bc + ac + ab + bc + ac}} + \dfrac{{2007}}{{\dfrac{{{{\left( {a + b + c} \right)}^2}}}{3}}}\left( {BDT:Cauchy - Schwarz;BDT:\left( 1 \right)} \right)\\
= \dfrac{9}{{{{\left( {a + b + c} \right)}^2}}} + \dfrac{{6021}}{{{{\left( {a + b + c} \right)}^2}}}\\
= \dfrac{{6030}}{{{{\left( {a + b + c} \right)}^2}}}\\
\ge \dfrac{{6030}}{{{3^2}}}\left( {Do:a + b + c \le 3} \right)\\
= 670
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{{{a^2} + {b^2} + {c^2}}} = \dfrac{1}{{ab + bc + ac}}\\
a = b = c\\
a + b + c = 3
\end{array} \right.\\
\Leftrightarrow a = b = c = 1
\end{array}$
Ta có điều phải chứng minh
