Giải thích các bước giải:
\(\begin{array}{l}
a,\\
\frac{{{a^2}}}{{a + b}} + \frac{{{b^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}} = \frac{{16}}{{17}}\\
\Leftrightarrow \left( {a + b + c} \right) - \left( {\frac{{{a^2}}}{{a + b}} + \frac{{{b^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}}} \right) = a + b + c - \frac{{16}}{{17}}\\
\Leftrightarrow \left( {a - \frac{{{a^2}}}{{a + b}}} \right) + \left( {b - \frac{{{b^2}}}{{b + c}}} \right) + \left( {c - \frac{{{c^2}}}{{c + a}}} \right) = a + b + c - \frac{{16}}{{17}}\\
\Leftrightarrow \frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}} = a + b + c - \frac{{16}}{{17}}\\
\frac{{{{\left( {a + b} \right)}^2}}}{{a + b}} + \frac{{{{\left( {b + c} \right)}^2}}}{{b + c}} + \frac{{{{\left( {c + a} \right)}^2}}}{{c + a}} = 2\left( {a + b + c} \right)\\
\Leftrightarrow \frac{{{a^2} + 2ab + {b^2}}}{{a + b}} + \frac{{{b^2} + 2bc + {c^2}}}{{b + c}} + \frac{{{c^2} + 2ca + {a^2}}}{{c + a}} = 2\left( {a + b + c} \right)\\
\Leftrightarrow \left( {\frac{{{a^2}}}{{a + b}} + \frac{{{b^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}}} \right) + 2\left( {\frac{{ab}}{{a + b}} + \frac{{bc}}{{b + c}} + \frac{{ca}}{{c + a}}} \right) + \left( {\frac{{{b^2}}}{{a + b}} + \frac{{{c^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}}} \right) = 2\left( {a + b + c} \right)\\
\Leftrightarrow \frac{{16}}{{17}} + 2.\left( {a + b + c - \frac{{16}}{{17}}} \right) + \left( {\frac{{{b^2}}}{{a + b}} + \frac{{{c^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}}} \right) = 2\left( {a + b + c} \right)\\
\Leftrightarrow \left( {\frac{{{b^2}}}{{a + b}} + \frac{{{c^2}}}{{b + c}} + \frac{{{c^2}}}{{c + a}}} \right) = \frac{{16}}{{17}}
\end{array}\)
b,
Áp dụng Bất Đẳng Thức Cô - si ta có:
\(\begin{array}{l}
M = 4{x^2} - 3x + \frac{1}{{4x}} + 2017\\
= \left( {4{x^2} - 4x + 1} \right) + \left( {x + \frac{1}{{4x}}} \right) + 2016\\
= {\left( {2x - 1} \right)^2} + \left( {x + \frac{1}{{4x}}} \right) + 2016\\
\ge 0 + 2\sqrt {x.\frac{1}{{4x}}} + 2016 = 2.\frac{1}{2} + 2016 = 2017
\end{array}\)
Dấu '=' xảy ra khi và chỉ khi \(\left\{ \begin{array}{l}
{\left( {2x - 1} \right)^2} = 0\\
x = \frac{1}{{4x}}
\end{array} \right. \Leftrightarrow x = \frac{1}{2}\)
