Giúp mình tí rồi

  • Thread starter coganghoctapthatgioi
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vansang02121998

$P=\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(c+a)}+\dfrac{1}{c^3(a+b)}$

\[\large P=\dfrac{b^2 c^2}{ab+ac}+\dfrac{a^2 c^2}{ab+bc}+\dfrac{a^2b^2}{ac+bc} \geq \dfrac{(ab+ac+bc)^2}{2(ab+ac+bc)}=\dfrac{ab+ac+bc}{2} \geq \dfrac{3\sqrt[3]{a^2b^2c^2}}{2} = \dfrac{3}{2}\]
 
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