IIT-JEE 1983Question 1
Show that $1+x\ln\left(x+\sqrt{x^{2}+1}\right)\ge\sqrt{1+x^{2}}$ for all $x\ge0$.
Differentiation and its Application
Answerproved
Solution
Concept. Let $h(x)=1+x\ln\left(x+\sqrt{x^2+1}\right)-\sqrt{1+x^2}$. Show $h'\ge0$ and use $h(0)=0$. Formulas used. (1) $\dfrac{d}{dx}\ln\left(x+\sqrt{x^2+1}\right)=\dfrac{1}{\sqrt{x^2+1}}$Step 1 - Derivative. By (1), $$h'(x)=\ln\left(x+\sqrt{x^2+1}\right)+\frac{x}{\sqrt{x^2+1}}-\frac{x}{\sqrt{1+x^2}}=\ln\left(x+\sqrt{x^2+1}\right).$$ Step 2 - Sign. For $x\ge0$, $x+\sqrt{x^2+1}\ge1$, so $h'(x)\ge0$. Step 3 - Conclude. $h$ is non-decreasing on $[0,\infty)$ and $h(0)=1-1=0$, so $h(x)\ge0$. Hence proved.