Solve the inequality (h + 0.9)/√(1 + h²) < h.

Step-by-step solution

1. 1

   Clear the denominator safely. Since $\sqrt{1+h^2} \ge 1 > 0$ for every real $h$, multiplying through does not flip the inequality:

   $h + 0.9 < h\sqrt{1+h^2}$

   Then isolate the $0.9$ by subtracting $h$:

   $0.9 < h\left(\sqrt{1+h^2} - 1\right)$

2. 2

   Eliminate h ≤ 0 immediately. The bracket satisfies $\sqrt{1+h^2} - 1 \ge 0$ for all $h$, so if $h \le 0$ the right side is $\le 0$, while the left side is $0.9 > 0$. No non-positive $h$ can work, and the search reduces to $h > 0$.

3. 3

   Square with both sides known positive. For $h > 0$ divide by $h$ and add $1$:

   $1 + \frac{0.9}{h} < \sqrt{1+h^2}$

   Both sides are now positive, so squaring is an equivalence rather than a one-way implication:

   $1 + \frac{1.8}{h} + \frac{0.81}{h^2} < 1 + h^2$

4. 4

   Clear the powers of h to get a quartic. Subtract $1$ and multiply by $h^2 > 0$:

   $1.8h + 0.81 < h^4 \quad \Longrightarrow \quad g(h) = h^4 - 1.8h - 0.81 > 0$

5. 5

   Find the single positive root of the quartic. On $h > 0$, $g'(h) = 4h^3 - 1.8$ vanishes only at $h = \sqrt[3]{0.45} \approx 0.7663$, so $g$ falls then rises: with $g(0) = -0.81 < 0$ there is exactly one positive root. Newton's method from $h_0 = 1.34$ gives

   $h_* = 1.3397213599 \quad (g(h_*) \approx 10^{-39} \ \text{at 40-digit precision})$

   The often-quoted value $1.35082$ is wrong: there $g = +0.0881$, well clear of zero.

6. 6

   Assemble the solution set and check it. $g > 0$ for $h > h_*$, so

   $h > 1.3397213599\ldots$

   Direct substitution into the original inequality confirms the boundary: at $h = 1.339$ the left side minus the right side is $+0.00075$ (fails), at $h = 1.3398$ it is $-0.000082$ (holds), and at $h = 1.35$ it is $-0.0107$ (holds). Rounded, the answer is $h > 1.3397$.