EXERCISE: Cho a,b,c > 0. Chứng minh rằng
$\sqrt {\frac{{{{\left( {a + b} \right)}^3}}}{{8ab\left( {4a + 4b + c} \right)}}} + \sqrt {\frac{{{{\left( {b + c} \right)}^3}}}{{8bc\left( {4b + 4c + a} \right)}}} + \sqrt {\frac{{{{\left( {c + a} \right)}^3}}}{{8ca\left( {4c + 4a + b} \right)}}} \ge 1$
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$\sum {\sqrt {\frac{{{{\left( {a + b} \right)}^3}}}{{8ab\left( {4a + 4b + c} \right)}}} } \ge 1$
Started By NLT, 15-05-2012 - 10:11
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