1708$Here expression is $f(x)=\frac{(x+a)(x+b)}{(x+c)}$\\\\ Let $(x+c)=t$, Then $x=t-c$\\\\ So $(x+a)=(t+a-c)$ and $(x+b)=(t+b-c)$\\\\ So $f(t)=\frac{(t+a-c)(t+b-c)}{t}=\frac{t^2+t\left \{ (a-c)+(b-c) \right \}+(a-c).(b-c)}{t}$\\\\ So $f(t)=t+\frac{(a-c).(b-c)}{t}+\left \{ (a-c)+(b-c) \right \}\geq 2.\sqrt{(a-c).(b-c)}+(a-c)+(b-c)$\\\\ Here We Have Use $A.M\geq G.M$\\\\ So $f(t)\geq (\sqrt{a-c})^2+(\sqrt{b-c})^2+2.\sqrt{(a-c).(b-c)}$\\\\ So $\boxed{\boxed{f(t)\geq \left \{\sqrt{a-c}+\sqrt{b-c} \right \}^2}}$
