[Toán] $abc + \frac{1}{ab + bc + ca} \geq \frac{4}{3}$

E

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$abc+\dfrac{1}{ab+bc+ca}=\dfrac{a+b+c}{3}+\dfrac{1}{ab+bc+ca} \ge \dfrac{\sqrt{3}.\sqrt{ab+bc+ca}}{3}+\dfrac{1}{ab+bc+ca}=\dfrac{\sqrt{3}.\sqrt{ab+bc+ca}}{9}+\dfrac{\sqrt{3}.\sqrt{ab+bc+ca}}{9}+\dfrac{1}{ab+bc+ca}+\dfrac{\sqrt{3}.\sqrt{ab+bc+ca}}{9} \ge 3.\dfrac{1}{3}+\dfrac{1}{3}=\dfrac{4}{3}$
 
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