lean_workbook_plus_6174

PP. Let $a$ , $b$ , $c \in R^{+}$ . Solve the equation $\sqrt{a+bx}+\sqrt{b+cx}+\sqrt{c+ax}=\sqrt{b-ax}+\sqrt{c-bx}+\sqrt{a-cx}$ .\n\nProof. $f(x)=\left(\sqrt{a+bx}+\sqrt{b+cx}+\sqrt{c+ax}\right)-\left(\sqrt{b-ax}+\sqrt{c-bx}+\sqrt{a-cx}\right)$ is (strict) increasing $\left(\nearrow\right)$ . Equation\n\nbecomes $f(x)=0$ , where $f$ is injectively. Hence $f(x)=0$ has at most one zero. Since $f(0)=0$ obtain $x=0$ is alone zero.