$\boxed{\text{Bài toán}}$
Cho $x,y,z> 0$ thỏa $27xyz=1$.CMR:
$\sum \frac{(xy)^{2}}{(xz+y^{2})^{2}}\geq \frac{1}{108(x^{2}z+y^{2}x+z^{2}y)^{2}}$
$\boxed{\text{Duy Thai2002}}$
Đặt $a = 3x,b = 3y,c = 3z$ suy ra $abc=1$
$BDT \Leftrightarrow \sum {\frac{{{{(ab)}^2}}}{{{{({b^2} + ca)}^2}}}} \ge \frac{{27}}{{4{{\left( {\sum {a{b^2}} } \right)}^2}}}$
Giải:
$\sum {\frac{{{{(ab)}^2}}}{{{{({b^2} + ca)}^2}}}} = \sum {\frac{{{{({a^2}b)}^2}}}{{{{(a{b^2} + c{a^2})}^2}}}} \ge \frac{1}{3}{\left( {\sum {\frac{{{a^2}b}}{{a{b^2} + c{a^2}}}} } \right)^2} \ge \frac{1}{3}{\left( {\frac{{{{\left( {\sum {a\sqrt b } } \right)}^2}}}{{2\sum {a{b^2}} }}} \right)^2} \ge \frac{1}{3}\frac{{{{\left( {3\sqrt[3]{{abc\sqrt {abc} }})} \right)}^4}}}{{4{{\left( {\sum {a{b^2}} } \right)}^2}}} = \frac{{27}}{{4{{\left( {\sum {a{b^2}} } \right)}^2}}}$
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