Inequality involving exponential function

Question

It is trivial to show that $$\sqrt {1+2x} < e^x, \quad \text{for}\, x>0$$ Is it true the stronger inequality $$\sqrt {1+\frac{8x}{4-x}} < e^x, \quad \text{for}\, x>0?$$

Answer 1

If $x\ge 0$ then $1+x+\frac12x^2\le e^x$ because the LHS is a truncation of the power series for $e^x$ and the missing terms are positive. Thus if $x\in[0,3]$ then $$ 4+7x \le 4+7x+2x^2(3-x) = (4-x)(1+2x+2x^2) \le (4-x)e^{2x} $$

Answer 2

For some $a<4$ this appears to be true.