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Đặt ($a;b;c$)→($\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}$).Khi đó điều kiện bài toán là

$2x+8y+21z \le 12xyz$

$2x+8y+21z \le12xyz$ \Rightarrow $3z \ge \dfrac{2x+8y}{4xy−7}$

\Rightarrow $P \ge x+2y+\dfrac{2x+8y}{4xy−7}=x+ \dfrac{11}{2x}+\dfrac{1}{2x}[(4xy−7)+\dfrac{4x^2+28}{4xy−7}]$

$ \ge x+\dfrac{11}{2x}+\dfrac{1}{x}.\sqrt{4x^2+28}=x+ \dfrac{11}{2x}+\dfrac{3}{2}.\sqrt{(1+\dfrac{7}{9})(1+\dfrac{7}{x^2})}$

$ \ge x+\dfrac{11}{2x}+\dfrac{3}{2}(1+\dfrac{7}{3x})=x+ \dfrac{9}{x}+\dfrac{3}{2} \ge \dfrac{15}{2}$

Nguồn:VMF
 
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