BDT rất rất khó!!!!!

C

conga222222

ahxd123 said:
\sqrt[2]{\frac{a^3}{(b+c)^3}}+\sqrt[2]{\frac{b^3}{(a+c)^3}} +\sqrt[n2]{\frac{c^3}{(a+b)^3}}.

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\[\begin{array}{l}
S = \sqrt[2]{{\frac{{{a^3}}}{{{{(b + c)}^3}}}}} + \sqrt[2]{{\frac{{{b^3}}}{{{{(a + c)}^3}}}}} + \sqrt[2]{{\frac{{{c^3}}}{{{{(a + b)}^3}}}}}\\
co.si:\\
\sqrt {\frac{{{a^3}}}{{{{\left( {b + c} \right)}^3}}}} + \frac{{\sqrt {\frac{a}{{b + c}}} }}{2} \ge 2\sqrt {\frac{{\sqrt {\frac{{{a^4}}}{{{{\left( {b + c} \right)}^4}}}} }}{2}} = \frac{{\sqrt 2 a}}{{b + c}} \to \sqrt {\frac{{{a^3}}}{{{{\left( {b + c} \right)}^3}}}} \ge \frac{{\sqrt 2 a}}{{b + c}} - \frac{1}{2}\sqrt {\frac{a}{{b + c}}} = \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\frac{a}{{b + c}} - \frac{1}{{4\sqrt 2 }} + \frac{a}{{2\sqrt 2 \left( {b + c} \right)}} + \frac{1}{{4\sqrt 2 }} - \frac{1}{2}\sqrt {\frac{a}{{b + c}}} \\
\frac{1}{{4\sqrt 2 }} + \frac{a}{{2\sqrt 2 \left( {b + c} \right)}} \ge 2\sqrt {\frac{a}{{16\left( {b + c} \right)}}} = \frac{1}{2}\sqrt {\frac{a}{{b + c}}} \to \frac{a}{{2\sqrt 2 \left( {b + c} \right)}} + \frac{1}{{4\sqrt 2 }} - \frac{1}{2}\sqrt {\frac{a}{{b + c}}} \ge 0\\
\to \sqrt {\frac{{{a^3}}}{{{{\left( {b + c} \right)}^3}}}} \ge \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\frac{a}{{b + c}} - \frac{1}{{4\sqrt 2 }}\\
tuong.tu\\
\sqrt {\frac{{{b^3}}}{{{{\left( {a + c} \right)}^3}}}} \ge \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\frac{b}{{a + c}} - \frac{1}{{4\sqrt 2 }}\\
\sqrt {\frac{{{c^3}}}{{{{\left( {b + a} \right)}^3}}}} \ge \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\frac{c}{{b + a}} - \frac{1}{{4\sqrt 2 }}\\
\to S \ge \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\left( {\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}} \right) - \frac{3}{{4\sqrt 2 }}\\
xet:p = \frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}\\
\to P + 3 = \frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}} + \frac{{a + b}}{{a + b}} + \frac{{b + c}}{{b + c}} + \frac{{a + c}}{{a + c}} = a\left( {\frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}}} \right) + b\left( {\frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}}} \right) + c\left( {\frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}}} \right)\\
bunhiacopski\\
\left( {\frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}}} \right)\left( {\left( {b + c} \right) + \left( {a + b} \right) + \left( {a + c} \right)} \right) \ge {\left( {1 + 1 + 1} \right)^2} = 9 \to \frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}} \ge \frac{9}{{2\left( {a + b + c} \right)}}\\
\to P + 3 \ge \frac{{9a}}{{2\left( {a + b + c} \right)}} + \frac{{9b}}{{2\left( {a + b + c} \right)}} + \frac{{9c}}{{2\left( {a + b + c} \right)}} = \frac{9}{2} \to P \ge \frac{3}{2}\\
\to S \ge \frac{3}{2}\left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right) - \frac{3}{{4\sqrt 2 }} = \frac{{3\sqrt 2 }}{2} - \frac{{3\sqrt 2 }}{8} - \frac{{3\sqrt 2 }}{8} = \frac{{3\sqrt 2 }}{4}\\
dau = \leftrightarrow a = b = c
\end{array}\]
 
M

magiciancandy

conga222222 said:
\[\begin{array}{l}
S = \sqrt[2]{{\frac{{{a^3}}}{{{{(b + c)}^3}}}}} + \sqrt[2]{{\frac{{{b^3}}}{{{{(a + c)}^3}}}}} + \sqrt[2]{{\frac{{{c^3}}}{{{{(a + b)}^3}}}}}\\
co.si:\\
\sqrt {\frac{{{a^3}}}{{{{\left( {b + c} \right)}^3}}}} + \frac{{\sqrt {\frac{a}{{b + c}}} }}{2} \ge 2\sqrt {\frac{{\sqrt {\frac{{{a^4}}}{{{{\left( {b + c} \right)}^4}}}} }}{2}} = \frac{{\sqrt 2 a}}{{b + c}} \to \sqrt {\frac{{{a^3}}}{{{{\left( {b + c} \right)}^3}}}} \ge \frac{{\sqrt 2 a}}{{b + c}} - \frac{1}{2}\sqrt {\frac{a}{{b + c}}} = \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\frac{a}{{b + c}} - \frac{1}{{4\sqrt 2 }} + \frac{a}{{2\sqrt 2 \left( {b + c} \right)}} + \frac{1}{{4\sqrt 2 }} - \frac{1}{2}\sqrt {\frac{a}{{b + c}}} \\
\frac{1}{{4\sqrt 2 }} + \frac{a}{{2\sqrt 2 \left( {b + c} \right)}} \ge 2\sqrt {\frac{a}{{16\left( {b + c} \right)}}} = \frac{1}{2}\sqrt {\frac{a}{{b + c}}} \to \frac{a}{{2\sqrt 2 \left( {b + c} \right)}} + \frac{1}{{4\sqrt 2 }} - \frac{1}{2}\sqrt {\frac{a}{{b + c}}} \ge 0\\
\to \sqrt {\frac{{{a^3}}}{{{{\left( {b + c} \right)}^3}}}} \ge \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\frac{a}{{b + c}} - \frac{1}{{4\sqrt 2 }}\\
tuong.tu\\
\sqrt {\frac{{{b^3}}}{{{{\left( {a + c} \right)}^3}}}} \ge \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\frac{b}{{a + c}} - \frac{1}{{4\sqrt 2 }}\\
\sqrt {\frac{{{c^3}}}{{{{\left( {b + a} \right)}^3}}}} \ge \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\frac{c}{{b + a}} - \frac{1}{{4\sqrt 2 }}\\
\to S \ge \left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right)\left( {\frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}} \right) - \frac{3}{{4\sqrt 2 }}\\
xet:p = \frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}}\\
\to P + 3 = \frac{a}{{b + c}} + \frac{b}{{c + a}} + \frac{c}{{a + b}} + \frac{{a + b}}{{a + b}} + \frac{{b + c}}{{b + c}} + \frac{{a + c}}{{a + c}} = a\left( {\frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}}} \right) + b\left( {\frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}}} \right) + c\left( {\frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}}} \right)\\
bunhiacopski\\
\left( {\frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}}} \right)\left( {\left( {b + c} \right) + \left( {a + b} \right) + \left( {a + c} \right)} \right) \ge {\left( {1 + 1 + 1} \right)^2} = 9 \to \frac{1}{{b + c}} + \frac{1}{{a + b}} + \frac{1}{{a + c}} \ge \frac{9}{{2\left( {a + b + c} \right)}}\\
\to P + 3 \ge \frac{{9a}}{{2\left( {a + b + c} \right)}} + \frac{{9b}}{{2\left( {a + b + c} \right)}} + \frac{{9c}}{{2\left( {a + b + c} \right)}} = \frac{9}{2} \to P \ge \frac{3}{2}\\
\to S \ge \frac{3}{2}\left( {\sqrt 2 - \frac{1}{{2\sqrt 2 }}} \right) - \frac{3}{{4\sqrt 2 }} = \frac{{3\sqrt 2 }}{2} - \frac{{3\sqrt 2 }}{8} - \frac{{3\sqrt 2 }}{8} = \frac{{3\sqrt 2 }}{4}\\
dau = \leftrightarrow a = b = c
\end{array}\]
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wawawa,siêu wá@-)............................................................................................
 
V

vodichhocmai

[TEX] \huge \blue \sqrt[]{\frac{a^3}{(b+c)^3}} \ge \frac{6\frac{a}{b+c} -1}{4\sqrt{2}[/TEX]
 
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