[Toán 12] Tìm GTNN của biểu thức

J

jet_nguyen

Giải :
Ta có :
$$P + 5 = \dfrac{3(b + c)}{2a} + \dfrac{4a + 3c}{3b} + \left (\dfrac{12(b - c)}{2a + 3c} + 4\right ) + 1 $$$$= \left (\dfrac{3b}{2a} + \dfrac{2a}{3b} \right )+ \left (\dfrac{3c}{2a} + \dfrac{3c}{3b} \right )+ \left (\dfrac{2a}{2a} + \dfrac{2a}{3b}\right ) + \dfrac{4(3b + 2a)}{2a + 3c}$$ $$ \ge 2 + \dfrac{4.2a}{2a + 3b} + \dfrac{4.3c}{3b + 2a} + \dfrac{4(3b + 2a)}{2a + 3c}$$$$ = 2 + 4\left (\dfrac{2a + 3c}{3b + 2a} + \dfrac{3b + 2a}{2a + 3c}\right ) \ge 2 + 4.2 = 10$$$\Longrightarrow P_{min} = 5$. Khi và chỉ khi: $ 2a = 3b = 3c$


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