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tim GTNN cua: \frac{2}{a^2+b^2}+\frac{35}{ab}+2ab
biet:a,b>0. a+b=4.
Anh chưa nghĩ được cách nào hay hơn em à:
\[\begin{array}{l}
P = \frac{2}{{{a^2} + {b^2}}} + \frac{{35}}{{ab}} + 2ab\\
a + b = 4\\
\frac{{35}}{{ab}} + 2ab \ge 2\sqrt {\frac{{35}}{{ab}}.2ab} = 2\sqrt {70} \\
Q = \sqrt {\frac{{{a^2} + {b^2}}}{2}} \\
Q \ge A \Rightarrow \sqrt {\frac{{{a^2} + {b^2}}}{2}} \ge \frac{{a + b}}{2} \Leftrightarrow \frac{{{a^2} + {b^2}}}{2} \ge 4 \Leftrightarrow \frac{2}{{{a^2} + {b^2}}} \ge \frac{1}{4}\\
\Rightarrow \frac{2}{{{a^2} + {b^2}}} + \frac{{35}}{{ab}} + 2ab \ge 2\sqrt {70} + \frac{1}{4} \Leftrightarrow A \ge \frac{{8\sqrt {70} + 1}}{4}\\
dau - bang \Leftrightarrow \left\{ \begin{array}{l}
\frac{{35}}{{ab}} = 2ab\\
a = b
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2{a^4} = 35\\
a = b
\end{array} \right. \Leftrightarrow {a^4} = {b^4} = \frac{{35}}{2}
\end{array}\]
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