$Show that 1 + x \log_{e} (x + \sqrt{(x^{2} + 1))} \geq \sqrt{(1 + x^{2}}) for all x \geq 0$

Moderate

Solution

$\begin{array}{l} Let f (x) = 1 + x \log_{e} (x + \sqrt{(x^{2} + 1))} - \sqrt{(1 + x^{2}}) \\ \Rightarrow f^{'} (x) = \ln (x + \sqrt{(x^{2} + 1)} + \frac{x}{(x + \sqrt{x^{2} + 1)})} \\ (1 + \frac{2 x}{2 \sqrt{x^{2} + 1})}) - \frac{2 x}{2 \sqrt{(1 + x^{2}})} \\ \Rightarrow f^{'} (x) = 1 n (x + \sqrt{x^{2} + 1)} \geq 0 \forall x \geq 0 \\ so, f (x) increases when x \geq 0 \\ hence x \geq 0 \Rightarrow f (x) \geq f (0) \\ Thus, 1 + x \ln (x + \sqrt{x^{2} + 1)}) - \sqrt{(1 + x^{2})} \geq 0 for x \geq 0 \\ \begin{array}{cl} So, & 1 + x \ln (x + \sqrt{x^{2} + 1}) \geq \sqrt{(1 + x^{2})} for x \geq 0 \end{array} \end{array}$

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