IIT-JEE 1979Question 1
Prove that the minimum value of $\dfrac{(a+x)(b+x)}{c+x}$, $a,b>c$, $x>-c$, is $\left(\sqrt{a-c}+\sqrt{b-c}\right)^{2}$.
Differentiation and its Application
Answerproved
Solution
Concept. Put $t=c+x>0$; the expression becomes $t+\frac{pq}{t}+p+q$, which AM-GM bounds. Formulas used. (1) AM-GM: $t+\frac{k}{t}\ge2\sqrt k$ for $t,k>0$, equality at $t=\sqrt k$Step 1 - Substitute. $t=c+x>0$, $p=a-c>0$, $q=b-c>0$. Then $a+x=t+p$, $b+x=t+q$: $$\frac{(a+x)(b+x)}{c+x}=\frac{(t+p)(t+q)}{t}=t+\frac{pq}{t}+p+q.$$ Step 2 - Bound. By (1), $$t+\frac{pq}{t}+p+q\ge2\sqrt{pq}+p+q=\left(\sqrt p+\sqrt q\right)^2.$$ Step 3 - Attained. Equality at $t=\sqrt{pq}$, i.e. $x=\sqrt{(a-c)(b-c)}-c>-c$. So the minimum is $\left(\sqrt{a-c}+\sqrt{b-c}\right)^2$. Hence proved.