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The set of all x for which the inequality 1 + xln(1+(1+x2) ≥ √(1+x2)

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\(f(n) = 1 + xln(x + \sqrt{x^2 + 1}) - \sqrt{1 + x^2}\)

\(f'(x) = ln (x + \sqrt{x^2 +1}) + \frac x{x + \sqrt{x^2 + 1}} \times \left[1 + \frac x{\sqrt{x^2 + 1}}\right] - \frac x{\sqrt{1 + x^2}}\)

\(= ln (x + \sqrt{x^2 +1}) + \frac x{ \sqrt{x^2 + 1}} - \frac x{\sqrt{1 + x^2}}\)

\(= ln (x + \sqrt{x^2 +1}) \)

\(f'(x) = ln (x + \sqrt{x^2 + 1})\)    \(x \ge 0\)

\(f'(x) \ge 0\)

f(x) is an \(f^n \;\forall \;x \ge 0\)

⇒ \(\;\forall \;x \ge 0\)

\(f(x) \ge f(0)\)

⇒ \(1 + x ln(x + \sqrt{x^2 + 1}) - \sqrt{1 + x^2} \ge 0\)

⇒ \(1 + x ln(x + \sqrt{x^2 + 1}) \ge \sqrt{1 + x^2}\) \(\;\forall \;x \ge 0\) 

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