Bài thiếu ĐK : a ; b ; c > 0
Ta có : P = \sum \frac{3+a^2}{b+c} = \sum \frac{3}{b+c} + \sum \frac{a^2}{b+c} = 3.\sum \frac{1}{b+c} + \sum \frac{a^2}{b+c}
Vì a ; b ; c > 0 nên AD BĐT Cauchy - Schwarz ta được :
\sum \frac{1}{b+c} \geq \frac{9}{2(a+b+c)} = \frac{9}{6} = 3/2 ( vì a + b + c = 3 )
\Rightarrow 3.\sum \frac{1}{b+c} \geq \frac{9}{2}
\sum \frac{a^2}{b+c} \geq \frac{(a+b+c)^2}{2(a+b+c)} = \frac{a+b+c}{2} = \frac{3}{2}
( vì a + b + c = 3 )
Suy ra : P \geq \frac{9}{2} + \frac{3}{2} = 6
Dấu " = " xảy ra <=> a = b = c = 1 (t/m)
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