I am given $$2x = a-\frac{1}{a} \;\text{and} \; 2y=b-\frac{1}{b} .$$ I have to find the value of $ xy +\sqrt{(x²+1)(y²+1)}.$
One way to solve this is to simply putting the value of $x$ and $y$ and get the answer, but that's a very lengthy process though lucid. But I am looking for a short cut method or some kind of tricks to solve the problem.
If anyone can visualize it, I wil be grateful for sharing. Thank you.
I do not think there is any special trick in solving the problem. $$x=\frac{a}{2}-\frac{1}{2a}$$ $$y=\frac{b}{2}-\frac{1}{2b}$$ $$xy=\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab}$$ $$x^2+1 = \frac{a^2}{4}-\frac{1}{2}+\frac{1}{4a^2}+1=\frac{a^2}{4}+\frac{1}{2}+\frac{1}{4a^2}=(\frac{a}{2}+\frac{1}{2a})^2$$ $$y^2+1 = \frac{b^2}{4}-\frac{1}{2}+\frac{1}{4b^2}+1=\frac{b^2}{4}+\frac{1}{2}+\frac{1}{4b^2}=(\frac{b}{2}+\frac{1}{2b})^2$$ Putting in the above values $$xy+\sqrt{(x^2+1)(y^2+1)}=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+\sqrt{(\frac{a}{2}+\frac{1}{2a})^2*(\frac{b}{2}+\frac{1}{2b})^2}$$ Assuming that $=\frac{a}{2}+\frac{1}{2a}>0$ and $\frac{b}{2}+\frac{1}{2b}>0$ $$=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+(\frac{a}{2}+\frac{1}{2a})*(\frac{b}{2}+\frac{1}{2b})$$ $$=(\frac{ab}{4}-\frac{a}{4b}-\frac{b}{4a}+\frac{1}{4ab})+(\frac{ab}{4}+\frac{a}{4b}+\frac{b}{4a}+\frac{1}{4ab})$$ $$=\frac{ab}{4}+\frac{1}{4ab}+\frac{ab}{4}+\frac{1}{4ab}$$ So, $$xy+\sqrt{(x^2+1)(y^2+1)}=(\frac{ab}{2}+\frac{1}{2ab})$$
Take $t_1=a-\frac1a$ and $t_2=b-\frac1b$ and then solve
$$x=\frac{t_1}{2},y=\frac{t_2}{2}$$ Now $$xy+\sqrt{(x^2+1)(y^2+1)}$$ $$\left(\frac{t_1}{2}\right)\left(\frac{t_2}{2}\right)+\sqrt{\left(\left(\frac{t_1}{2}\right)^2+1\right)\left(\left(\frac{t_2}{2}\right)^2+1\right)}$$ $$=\frac{t_1t_2}{4}+\sqrt{\frac{t_1^2+4}{4}\cdot \frac{t_2^2+4}{4}}$$ $$=\frac{t_1t_2}{4}+\frac{\sqrt{(t_1^2+4)(t_2^2+4)}}{4}$$ and then substitute back $t_1=a-\frac1a,t_2=b-\frac1b$
