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Chứng minh rằng: $a+\dfrac{4}{(a-b)(b+1)^{2}}\geq 3$


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#1
cvp

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Cho $a>b\geq 0$.
Chứng minh rằng: $a+\dfrac{4}{(a-b)(b+1)^{2}}\geq 3$

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#2
Ispectorgadget

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$VT=(a-b)+\dfrac{b+1}{2}+\dfrac{b+1}{2}-1+\dfrac{4}{(a-b)(b+1)^2}\geq 4\sqrt[3]{(a-b).\dfrac{(b-1)^2}{4}.\dfrac{4}{(a-b)(b+1)^2}}-1=3$

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