Với a,b,c>0.abc=1.CMR 1/[a^3(b+c)] + 1/[b^3 (c+a)] + 1/ [c^3 (a+b)] >=3/2
[tex]\frac{1}{a^3.(b+c)}+\frac{1}{b^3.(c+a)}+\frac{1}{c^3.(a+b)}\\\\ =\frac{\frac{1}{a^2}}{a.(b+c)}+\frac{\frac{1}{b^2}}{b.(c+a)}+\frac{\frac{1}{c^2}}{c.(a+b)}[/tex]
áp dụng BĐT Bunhiacopxki dạng phân thức có:
[tex]\frac{\frac{1}{a^2}}{a.(b+c)}+\frac{\frac{1}{b^2}}{b.(c+a)}+\frac{\frac{1}{c^2}}{c.(a+b)}\geq \frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{a.(c+b)+b.(c+a)+c.(a+b)}\\\\ =\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2.(ab+bc+ca)}=\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2.(\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc})}=\frac{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2}{2.(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}\\\\ =\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\geq \frac{3\sqrt[3]{\frac{1}{abc}}}{2}=\frac{3}{2} (abc=1)[/tex]
dấu "=" xảy ra khi a=b=c và abc=1
=> a=b=c=1
vậy....
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