Bài toán. Cho p,q>0p+q=1. Chứng minh  rằng \dfrac{1}{p} + \dfrac{1}{q} \geq 4{\left( {p + \dfrac{1}{p}} \right)^2} + {\left( {q + \dfrac{1}{q}} \right)^2} \ge \frac{{25}}{2}.

Chứng minh. (CLAUDI AL S INA và ROGER B. NE L SON)

Ta có 1\ge 4pq\Rightarrow \dfrac{1}{p} + \dfrac{1}{q} \geq 4.

{2\left( {p + \dfrac{1}{p}} \right)^2} + {2\left( {q + \dfrac{1}{q}} \right)^2} \ge {\left( {p + \dfrac{1}{p} + q + \dfrac{1}{q}} \right)^2} \ge {\left( {1 + 4} \right)^2} = 25.

Theo Mathematics Magazine

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