[Toán 8]

L

linhhuyenvuong

biết a,b,c>0
cm: [TEX]\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}[/TEX]
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Ta có
A+3=[tex](\frac{a}{b+c}+1)+(\frac{b}{a+c}+1)+(\frac{c}{a+b}+1)[/tex]
=[tex] (a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})[/tex]
=[tex] \frac{1}{2} .[(a+b)+(a+c)+(b+c)](\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})[/tex] \geq[tex]\frac{1}{2} .9=\frac{9}{2}[/tex]
\Rightarrow [tex]A\geq\frac{3}{2}[/tex]
 
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