với a,b,c là các số dương thoả mãn $a+b+c \le 2$
cmr
$\sqrt{a^2+1/b^2} +\sqrt{b^2+1/c^2} +\sqrt{c^2+1/a^2} \ge \sqrt{97}/2$
$\begin{array}{l}
bunhiacopski:\\
\sqrt {\left( {{a^2} + \frac{1}{{{b^2}}}} \right)\left( {\frac{4}{9} + \frac{9}{4}} \right)} \ge \frac{{2a}}{3} + \frac{3}{{2b}}\\
tuong\;tu\;cho\;2\;bieu\;thuc\;sau\;roi\;cong\;lai\\
\to \frac{{\sqrt {97} }}{6}*VT \ge \frac{{2\left( {a + b + c} \right)}}{3} + \frac{3}{2}\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\\
\cos i:\\
\frac{{2a}}{3} + \frac{{2b}}{3} + \frac{{2c}}{3} + \frac{8}{{27a}} + \frac{8}{{27b}} + \frac{8}{{27b}} \ge 6\sqrt[6]{{\frac{{{2^3}*{8^3}}}{{{3^3}*{{27}^3}}}}} = \frac{{24}}{9}\\
\frac{{65}}{{54}}\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) \ge \frac{{65}}{{54}}*3*\sqrt[3]{{\frac{1}{{abc}}}} = \frac{{65}}{{18\sqrt[3]{{abc}}}} \ge \frac{{65}}{{18\left( {\frac{{a + b + c}}{3}} \right)}} \ge \frac{{65}}{{12}}\\
\to \frac{{\sqrt {97} }}{6}*VT \ge \frac{{24}}{9} + \frac{{65}}{{12}} = \frac{{97}}{{12}}\\
\leftrightarrow VT \ge \frac{{\sqrt {97} }}{2}\\
dau = \leftrightarrow a = b = c = \frac{2}{3}
\end{array}$
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