Prove that x²+1 > e^(-x²) for all x ≠0
Prove that $x^{2}+1\gt e^{-x^{2}}$ for all $x\neq0$.
Explanation of Method
One substitution into the tangent-line inequality.
Step 1: Start from Problem 453
There it was shown that $e^{t}\gt 1+t$ for every $t\neq0$. For $x\neq0$ the number $t = x^{2}$ is positive, hence non-zero, so
\[ e^{x^{2}}\gt 1+x^{2}. \tag{1} \]
Step 2: Invert
Both sides of (1) are positive, so taking reciprocals reverses the inequality: \[e^{-x^{2}} \lt \frac{1}{1+x^{2}}. \tag{2}\]
Step 3: Compare with $1+x^{2}$
Since $1+x^{2}\gt 1$ for $x\neq0$, we also have $\dfrac{1}{1+x^{2}}\lt 1\lt 1+x^{2}$. Chaining this with (2): $e^{-x^{2}} \lt \frac{1}{1+x^{2}} \lt 1+x^{2}$.
Step 4: Verification
\[x = 0:\ 1 = 1 \;\;\checkmark\ \text{(equality, so the hypothesis } x\neq0 \text{ is necessary)}\]
\[x = 0.5:\ 1.25 \gt e^{-0.25} = 0.7788008; \qquad x = 2:\ 5 \gt e^{-4} = 0.0183156 \;\;\checkmark\]
\[\text{The intermediate bound is sharper: } e^{-0.25} = 0.7788 \lt \frac{1}{1.25} = 0.8 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,x^{2}+1>e^{-x^{2}} \quad\text{for all } x\neq0, \qquad\text{with equality only at } x = 0\,}\]
Remark
Remark. Applying $e^{t}\gt 1+t$ with $t = -x^{2}$ instead gives $e^{-x^{2}}\gt 1-x^{2}$, a lower bound — useful, but the wrong direction for this problem. The right move is to bound $e^{x^{2}}$ from below and then invert, which also yields the sharper estimate $e^{-x^{2}}\lt \dfrac{1}{1+x^{2}}$ recorded above.