$\require{cancel}$I am pretty sure you can demonstrate this with contradiction. You can however prove this directly as follows: Let $\psi(x)$ be a real and normalized eigenfunction of the Hamiltonian $H$. It is a known theorem that the eigenvalue $E$ that corresponds to $\psi(x)$ is the expected value of $H$ on the eigenfunction itself. Thus, $$E = \langle H\rangle \Longleftrightarrow E = \int_{\mathbb{R}} \psi(x) \left( -\frac{\hbar^2}{2m} \frac{\text{d}^2}{\text{d}x^2} + V(x) \right) \psi(x) \, dx \Longrightarrow $$ $$E = - \frac{\hbar^2}{2m} \int_{\mathbb{R}} \psi(x) \psi''(x) \, dx + \int_{\mathbb{R}} V(x)\psi^2(x) \, dx$$ By using integration by parts on the first integral, we obtain: $$E = -\frac{\hbar^2}{2m} \left( \bcancel{\left[ \psi'(x) \psi(x) \right]{-\infty}^{+\infty}} - \int{\mathbb{R}} \left(\frac{\text{d}^2 \psi}{\text{d}x^2} \right)^2 \, dx \right) + \int_{\mathbb{R}} V(x) \psi^2(x) \, dx \Longrightarrow $$ $$E = + \frac{\hbar^2}{2m} \int_{\mathbb{R}} \left(\frac{\text{d}^2 \psi}{\text{d}x^2} \right)^2 \, dx + \int_{\mathbb{R}} V(x) \psi^2(x) \, dx$$ Since $\psi''(x)^2 \geqslant 0$, we have: $$\frac{\hbar^2}{2m} \int_{\mathbb{R}} \left(\frac{\text{d}^2 \psi}{\text{d}x^2} \right)^2 \, dx \geqslant 0$$ and thus, $$E \geqslant \int_{\mathbb{R}} V(x) \psi^2(x) \, dx$$ Finally, since $V(x)$ exhibits a minimum $\min V$, we have: $$V(x) \geqslant \min V \Longrightarrow V(x) \psi^2(x) \geqslant (\min V) \cdot \psi^2(x) \Longrightarrow $$ $$ \int_{\mathbb{R}} V(x) \psi^2(x) \, dx \geqslant \int_{\mathbb{R}} (\min V) \cdot \psi^2(x) = \min V \text{ since } \int_{\mathbb{R}} \psi^2(x) \, dx = 1$$ Consequently, $$E \geqslant \int_{\mathbb{R}} V(x) \psi^2(x) \, dx \geqslant \min V \Longrightarrow E \geqslant \min V $$ Now, there is an intuitive reason why this happens (that is, the eigenvalues are bounded below from the minimum value of the potential function). You very well demonstrated that the converse of the statement would imply $\psi(x)$ and $\psi''(x)$ to have the same sign in $\mathbb{R}$ (or perhaps zero, but it's not of interest here). 955
That means, that if $\psi(x)$ is positive, so is $\psi''(x)$. The second derivative of $\psi(x)$ however dictates the concavity of the $\psi(x)$. This condition renders $\psi(x)$ incapable of being normalizable in $\mathbb{R}$. You can perhaps see this by drawing a few graphs yourself. This is the argument <PERSON> persuades the reader to make in his book Introduction to Quantum Mechanics (The last exercise at the end of Chapter 2). 469