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B

braga

Ta có :

$gt\iff y+\sqrt{y^{2}+1}=2012(\sqrt{x^{2}+1}-x);x+\sqrt{x^{2}+1}=2012(\sqrt{y^{2}+1}-y) \\ \implies x+y+\sqrt{x^{2}+1}+\sqrt{y^{2}+1}=2012(\sqrt{x^{2}+1}+\sqrt{y^{2}+1})-2012(x+y) \\ \implies (x+y)(1+2012)=(2012-1)(\sqrt{x^{2}+1}+\sqrt{y^{2}+1})$
Đặt : $\sqrt{x^{2}+1}+\sqrt{y^{2}+1}=B$
$\implies P.2013=B.2011;B^{2}=x^{2}+y^{2}+2+2\sqrt{(x^{2}+1)(y^{2}+1)}\geq x^{2}+y^{2}+2+2(xy+1)=(x+y)^{2}+4=P^{2}+4 \\ \implies P^{2}.2013^{2}=B^{2}.2011^{2}\geq (P^{2}+4).2011^{2} \\ \implies \dfrac{P^{2}+4}{P^{2}}\leq \dfrac{2013^{2}}{2011^{2}} \\ \dfrac{4}{P^{2}}\leq \dfrac{2013^{2}-2011^{2}}{2011^{2}}=\dfrac{8048}{2011^{2}}\implies P\geq \dfrac{2011\sqrt{2012}}{2012}$
Vậy :
$MinP=\dfrac{2011\sqrt{2012}}{2012}\iff x=y=\dfrac{2011\sqrt{2012}}{4024}$
 
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