Cho ba số thực dương a, b, c thỏa mãn a+ b+ c = 3. Tìm GTNN của biểu thức
P = [TEX] \frac{b\sqrt{b}}{\sqrt{2a + b+ c}} +\frac{c\sqrt{c}}{\sqrt{2b + c + a}} + \frac{ a\sqrt{a}}{\sqrt{2c + a +b}} [/TEX]
\[\begin{array}{l}
P = \frac{{b\sqrt b }}{{\sqrt {2a + b + c} }} + \frac{{c\sqrt c }}{{\sqrt {2b + c + a} }} + \frac{{a\sqrt a }}{{\sqrt {2c + a + b} }}\\
a + b + c = 3\\
\to P = \frac{{b\sqrt b }}{{\sqrt {a + 3} }} + \frac{{c\sqrt c }}{{\sqrt {c + 3} }} + \frac{{a\sqrt a }}{{\sqrt {c + 3} }}\\
bunhiacopski:\\
36 = \left( {b + a + c} \right)\left( {\left( {a + 3} \right) + \left( {b + 3} \right) + \left( {c + 3} \right)} \right) \ge {\left( {\sqrt {b\left( {a + 3} \right)} + \sqrt {c\left( {b + 3} \right)} + \sqrt {a\left( {c + 3} \right)} } \right)^2}\\
\to \left( {\sqrt {b\left( {a + 3} \right)} + \sqrt {c\left( {b + 3} \right)} + \sqrt {a\left( {c + 3} \right)} } \right) \le 6\\
bunhiacopski\\
\to \left( {\frac{{b\sqrt b }}{{\sqrt {a + 3} }} + \frac{{c\sqrt c }}{{\sqrt {b + 3} }} + \frac{{a\sqrt a }}{{\sqrt {c + 3} }}} \right)\left( {\frac{{\sqrt {b\left( {a + 3} \right)} }}{4} + \frac{{\sqrt {c\left( {b + 3} \right)} }}{4} + \frac{{\sqrt {a\left( {c + 3} \right)} }}{4}} \right) \ge {\left( {\frac{{b + c + a}}{2}} \right)^2} = \frac{9}{4}\\
\to P \ge \frac{9}{{4*\left( {\frac{{\sqrt {b\left( {a + 3} \right)} + \sqrt {c\left( {b + 3} \right)} + \sqrt {a\left( {c + 3} \right)} }}{4}} \right)}} = \frac{9}{6} = \frac{3}{2}\\
dau = \leftrightarrow a = b = c = 1
\end{array}\]
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