Giải thích các bước giải:

Ta có:

Bài 1:

Ta có bất đẳng thức sau:

\(\begin{array}{l}
{\left( {a - b} \right)^2} \ge 0,\,\,\,\forall a,b\\
 \Leftrightarrow {a^2} + {b^2} \ge 2ab\\
 \Leftrightarrow {a^2} + {b^2} + 2ab \ge 4ab\\
 \Leftrightarrow {\left( {a + b} \right)^2} \ge 4ab\\
 \Leftrightarrow \dfrac{{a + b}}{{ab}} \ge \dfrac{4}{{a + b}}\\
 \Leftrightarrow \dfrac{1}{a} + \dfrac{1}{b} \ge \dfrac{4}{{a + b}}\,\,\,\,\,\,\,\,\,\,\,\left( * \right)
\end{array}\)

Dấu '=' xảy ra khi và chỉ khi \(a = b\)

Áp dụng bất đẳng thức trên ta được:

\(\begin{array}{l}
\dfrac{1}{{2a + b + c}} + \dfrac{1}{{a + 2b + c}} + \dfrac{1}{{a + b + 2c}}\\
 = \dfrac{1}{4}.\left( {\dfrac{4}{{\left( {a + b} \right) + \left( {a + c} \right)}} + \dfrac{4}{{\left( {a + b} \right) + \left( {b + c} \right)}} + \dfrac{4}{{\left( {a + c} \right) + \left( {b + c} \right)}}} \right)\\
 \le \dfrac{1}{4}.\left( {\dfrac{1}{{a + b}} + \dfrac{1}{{a + c}} + \dfrac{1}{{a + b}} + \dfrac{1}{{b + c}} + \dfrac{1}{{a + c}} + \dfrac{1}{{b + c}}} \right)\\
 \le \dfrac{1}{{16}}.\left( {\dfrac{4}{{a + b}} + \dfrac{4}{{a + c}} + \dfrac{4}{{a + b}} + \dfrac{4}{{b + c}} + \dfrac{4}{{a + c}} + \dfrac{4}{{b + c}}} \right)\\
 = \dfrac{1}{{16}}.\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{1}{c} + \dfrac{1}{b} + \dfrac{1}{c}} \right)\\
 = \dfrac{1}{{16}}.4.\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)\\
 = \dfrac{1}{4}.\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right) = \dfrac{1}{4}.2020 = 505
\end{array}\)

Dấu '=' xảy ra khi và chỉ khi \(a = b = c = \dfrac{{2020}}{3}\)

Vậy \(\dfrac{1}{{2a + b + c}} + \dfrac{1}{{a + 2b + c}} + \dfrac{1}{{a + b + 2c}} \le 505\)

Bài 2:

\(\begin{array}{l}
\dfrac{{{{\cos }^2}A + {{\cos }^2}B}}{{{{\sin }^2}A + {{\sin }^2}B}}\\
 = \dfrac{{\left( {1 - {{\sin }^2}A} \right) + \left( {1 - {{\sin }^2}B} \right)}}{{{{\sin }^2}A + {{\sin }^2}B}}\\
 = \dfrac{{2 - \left( {{{\sin }^2}A + {{\sin }^2}B} \right)}}{{{{\sin }^2}A + {{\sin }^2}B}}\\
 = \dfrac{2}{{{{\sin }^2}A + {{\sin }^2}B}} - 1\\
\dfrac{1}{2}\left( {{{\cot }^2}A + {{\cot }^2}B} \right)\\
 = \dfrac{1}{2}\left( {\dfrac{{{{\cos }^2}A}}{{{{\sin }^2}A}} + \dfrac{{{{\cos }^2}B}}{{{{\sin }^2}B}}} \right)\\
 = \dfrac{1}{2}\left( {\dfrac{{1 - {{\sin }^2}A}}{{{{\sin }^2}A}} + \dfrac{{1 - {{\sin }^2}B}}{{{{\sin }^2}B}}} \right)\\
 = \dfrac{1}{2}\left( {\dfrac{1}{{{{\sin }^2}A}} - 1 + \dfrac{1}{{{{\sin }^2}B}} - 1} \right)\\
 = \dfrac{1}{2}\left( {\dfrac{1}{{{{\sin }^2}A}} + \dfrac{1}{{{{\sin }^2}B}}} \right) - 1\\
 = \dfrac{1}{2}.\dfrac{{{{\sin }^2}A + {{\sin }^2}B}}{{{{\sin }^2}A.{{\sin }^2}B}} - 1\\
\dfrac{{{{\cos }^2}A + {{\cos }^2}B}}{{{{\sin }^2}A + {{\sin }^2}B}} = \dfrac{1}{2}\left( {{{\cot }^2}A + {{\cot }^2}B} \right)\\
 \Leftrightarrow \dfrac{2}{{{{\sin }^2}A + {{\sin }^2}B}} = \dfrac{1}{2}\dfrac{{{{\sin }^2}A + {{\sin }^2}B}}{{{{\sin }^2}A.{{\sin }^2}B}}\\
 \Leftrightarrow 4{\sin ^2}A.{\sin ^2}B = {\left( {{{\sin }^2}A + {{\sin }^2}B} \right)^2}\\
 \Leftrightarrow \left( {{{\sin }^2}A - {{\sin }^2}B} \right) = 0\\
 \Leftrightarrow {\sin ^2}A = {\sin ^2}B
\end{array}\)

Câu 2 bị thiếu dữ kiện rồi nhé!