\(x+\sqrt{1+x^2}=a\left(\sqrt{1+y^2}-y\right)\) (\(a=2015\))
\(\Leftrightarrow\sqrt{1+x^2}+ay=a\sqrt{1+y^2}-x\)
\(\Leftrightarrow1+x^2+a^2y^2+2ay\sqrt{1+x^2}=a^2+a^2y^2+x^2-2ax\sqrt{1+y^2}\)
\(\Leftrightarrow y\sqrt{1+x^2}+x\sqrt{1+y^2}=\frac{a^2-1}{2a}=b\)
\(\Leftrightarrow y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=b^2\)
\(\Leftrightarrow x^2y^2+x^2y^2+x^2+y^2+1+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=b^2+1\)
\(\Leftrightarrow x^2y^2+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}+\left(1+x^2\right)\left(1+y^2\right)=b^2+1\)
\(\Leftrightarrow\left(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\right)^2=b^2+1\)
\(\Leftrightarrow xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=\sqrt{b^2+1}\)
\(\Leftrightarrow xy+\frac{1+x^2+1+y^2}{2}\ge\sqrt{b^2+1}\)
\(\Leftrightarrow\left(x+y\right)^2\ge2\sqrt{b^2+1}-2\)
\(\Leftrightarrow x+y\ge\sqrt{2\sqrt{b^2+1}-2}\)
\(\Rightarrow P_{min}=\sqrt{2\sqrt{b^2+1}-2}\) khi \(x=y\)
Làm gọn kết quả lại:
\(b^2+1=\left(\frac{a^2-1}{2a}\right)^2+1=\frac{a^4+2a^2+1}{4a^2}=\left(\frac{a^2+1}{2a}\right)^2\)
\(\Rightarrow\sqrt{2\sqrt{b^2+1}-2}=\sqrt{\frac{a^2+1}{a}-2}=\sqrt{\frac{\left(a-1\right)^2}{a}}=\frac{a-1}{\sqrt{a}}=\frac{2014}{\sqrt{2015}}\)
Vậy \(P_{min}=\frac{2014}{\sqrt{2015}}\) khi \(x=y=\frac{1007}{\sqrt{2015}}\)
