Solve without hit and trial

$A=(a+\frac{1}{a})^2+(b+\frac{1}{b})^2$ $=a^2+\frac{1}{a^2}+b^2+\frac{1}{b^2}+4$ $=a^2+\frac{1}{16a^2}+b^2+\frac{1}{16b^2}+\frac{15}{16}(\frac{1}{a^2}+\frac{1}{b^2})+4$ We have $a^2+\frac{1}{16a^2}\geq\frac{1}{2}$ $b^2+\frac{1}{16b^2}\geq\frac{1}{2}$ $\frac{1}{a^2}+\frac{1}{b^2}\geq\frac{2}{ab}\geq 8$ because $(a+b)^2\geq 4ab\Leftrightarrow\frac{1}{ab}\geq 4$ Therefore $A\geq 1+4+\frac{15}{2}$ $\Leftrightarrow A\geq 12.5\Longrightarrow x=12.5$ The equality holds when $a=b=\frac{1}{2}$