Prove that sqrt(x) < 2x+ dfrac38 for all x > 0
Prove that $\sqrt{x} \lt 2x+\dfrac38$ for all $x\gt 0$.
Explanation of Method
The gap between the two sides never falls below ¼.
Step 1: Reformulate
Set
\[ f(x) = 2x+\frac38-\sqrt{x}, \qquad x\gt 0, \] so that the claim is $f(x)\gt 0$ throughout.
Step 2: Locate the minimum
\[ f'(x) = 2-\frac{1}{2\sqrt{x}} = 0 \;\Longleftrightarrow\; \sqrt{x} = \frac14 \;\Longleftrightarrow\; x = \frac{1}{16}, \]
\[ f''(x) = \frac{1}{4x^{3/2}} \gt 0 \quad\text{for all } x\gt 0, \] so $f$ is strictly convex and $x = \tfrac{1}{16}$ is its unique global minimum.
Step 3: The minimum value
\[ f\!\left(\frac{1}{16}\right) = \frac18+\frac38-\frac14 = \frac12-\frac14 = \frac14 \gt 0. \]
Step 4: Conclusion
Since the least value of $f$ is $\tfrac14\gt 0$, we have $f(x)\gt 0$ for every $x\gt 0$, that is
\[ \sqrt{x}\lt 2x+\frac38 \qquad (x\gt 0). \]
The inequality is strict everywhere, and cannot be improved by more than $\tfrac14$: at $x = \tfrac1{16}$ the two sides differ by exactly that amount.
Step 5: Verification
\[f(0.01) = 0.295; \qquad f\!\left(\tfrac1{16}\right) = 0.25; \qquad f(0.25) = 0.375; \qquad f(1) = 1.375 \;\;\checkmark\]
\[\text{Substituting } u = \sqrt{x}:\ 2u^{2}-u+\tfrac38 = 2\left(u-\tfrac14\right)^{2}+\tfrac14 \;\;\checkmark\]
\[\text{The completed square gives the same minimum } \tfrac14 \text{ at } u = \tfrac14,\ \text{i.e. } x = \tfrac1{16} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\sqrt{x} < 2x+\frac38 \quad (x>0), \qquad \min_{x>0}\left(2x+\frac38-\sqrt{x}\right) = \frac14 \text{ at } x = \frac1{16}\,}\]
Remark
Remark. The substitution in the verification turns the problem into a quadratic and disposes of it without calculus: the discriminant of $2u^{2}-u+\tfrac38$ is $1-3 = -2\lt 0$, so the expression never vanishes. The coefficient $\tfrac38$ is generous; the sharp constant for which $\sqrt{x}\le2x+c$ still holds is $c = \tfrac18$, where the parabola becomes tangent to the axis.