Show that $$1+x$$ $$In\left( {x + \sqrt {{x^2} + 1} } \right) \ge \sqrt {1 + {x^2}} $$ for all $$x \ge 0$$

Find the coordinates of the point on the curve $$y = {x \over {1 + {x^2}}}$$
where the tangent to the curve has the greatest slope.

If $$f(x)$$ and $$g(x)$$ are differentiable function for $$0 \le x \le 1$$ such that $$f(0)=2$$, $$g(0)=0$$, $$f(1)=6$$; $$g(1)=2$$, then show that there exist $$c$$ satisfying $$0 < c < 1$$ and $$f'(c)=2g'(c)$$.

If $$a{x^2} + {b \over x} \ge c$$ for all positive $$x$$ where $$a>0$$ and $$b>0$$ show that $$27a{b^2} \ge 4{c^3}$$.