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u la tr luoi go latex voi cai tieu de hong nha nhunh van dich duoc kiki

$\sum \frac{4a-1}{(2b+1)^{2}}$$\geq 1$

tương đương $\sum (\frac{4a-1}{(2b+1)^{2}}$+1)$\geq 4$

tương đương$\sum \frac{4a+4b+4b^{2}}{(2b+1)^{2}}$$\geq 4$

tương đương$\sum \frac{a+b+b^{2}}{(2b+1)^{2}}$$\geq 1$

theo bunhiacopski $(a+b+b^{2})(\frac{1}{a}+b+1)\geq (b+b+1)^{2}=(2b+1)^{2}$

suy ra $\sum \frac{a+b+b^{2}}{(2b+1)^{2}}\geq \frac{1}{\frac{1}{a}+b+1}=\frac{a}{ab+a+1}$

suy ra$\inline \sum \frac{a+b+b^{2}}{(2b+1)^2}\geq \frac{a}{ab+a+1}+\frac{b}{cb+b+1}+\frac{c}{ac+c+1}$

=$\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}$=1

Dấu "=" a=b=c=1.

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