Giá trị nhỏ nhất

E

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$P \ge \dfrac{4}{4a^2+4b^2+4}+\dfrac{2015}{ab}=\dfrac{1}{a^2+b^2+1}+\dfrac{2015}{ab}=\dfrac{1}{5-2ab}+\dfrac{2015}{ab}=\dfrac{1}{3(5-2ab)}+\dfrac{1}{3(5-2ab)}+\dfrac{1}{3(5-2ab)}+\dfrac{1}{9ab}+\dfrac{1}{9ab}+\dfrac{18133}{9ab} \ge \dfrac{25}{45}+\dfrac{18133}{9.1}=\dfrac{6046}{3}$

Dấu = xảy ra khi $a=b=1$
 
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