Cho ab>2019a+2020b. Cm a+b>(√2019+√2020) bình phuong

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
\left( {{a^2} + {x^2}} \right)\left( {{b^2} + {y^2}} \right)\\
 = {a^2}{b^2} + {a^2}{y^2} + {x^2}{b^2} + {x^2}{y^2}\\
 = \left( {{a^2}{b^2} + {x^2}{y^2}} \right) + \left( {{a^2}{y^2} + {x^2}{b^2}} \right)\\
 \ge \left( {{a^2}{b^2} + {x^2}{y^2}} \right) + 2ayxb = {\left( {ab + xy} \right)^2}\\
 \Rightarrow \left( {{a^2} + {b^2}} \right)\left( {{x^2} + {y^2}} \right) \ge {\left( {ab + xy} \right)^2}
\end{array}\)

\(\begin{array}{l}
ab > 2019a + 2020b\\
 \Leftrightarrow \frac{{2019a + 2020b}}{{ab}} < 1\\
 \Leftrightarrow \frac{{2019}}{b} + \frac{{2020}}{a} < 1
\end{array}\)

Áp dụng BĐT trên ta có:

\(\begin{array}{l}
\left( {\frac{{2019}}{b} + \frac{{2020}}{a}} \right)\left( {b + a} \right) \ge {\left( {\sqrt {\frac{{2019}}{b}.b}  + \sqrt {\frac{{2020}}{a}.a} } \right)^2} = {\left( {\sqrt {2019}  + \sqrt {2020} } \right)^2}\\
 \Rightarrow b + a \ge \frac{{{{\left( {\sqrt {2019}  + \sqrt {2020} } \right)}^2}}}{{\frac{{2019}}{b} + \frac{{2020}}{a}}} \ge {\left( {\sqrt {2019}  + \sqrt {2020} } \right)^2}
\end{array}\)