Cho $a>b\geq 0$.
Chứng minh rằng: $a+\dfrac{4}{(a-b)(b+1)^{2}}\geq 3$
Chứng minh rằng: $a+\dfrac{4}{(a-b)(b+1)^{2}}\geq 3$
Started By cvp, 23-12-2011 - 16:46
#1
Posted 23-12-2011 - 16:46
#2
Posted 23-12-2011 - 17:05
$VT=(a-b)+\dfrac{b+1}{2}+\dfrac{b+1}{2}-1+\dfrac{4}{(a-b)(b+1)^2}\geq 4\sqrt[3]{(a-b).\dfrac{(b-1)^2}{4}.\dfrac{4}{(a-b)(b+1)^2}}-1=3$
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