求证：二元实函数$1 + \frac{xy}{e^{\frac{1}{2}(x^2+y^2-2)}}$在$\mathbb{R^2}$上恒非负

Hey there! Let's work through this proof properly. Your initial approach hit a snag because taking the natural logarithm isn't valid when $xy$ is negative (logs aren't defined for negative numbers). Instead, we can split the problem into two clear cases based on the sign of $xy$, which makes the proof straightforward:

Case 1: $xy \ge 0$ #

When $xy$ is non-negative, the term $\frac{xy}{e^{{\frac{1}{2}(x}2+y^2-2)}}$ is also non-negative (since the exponential function is always positive). Adding this to 1 gives:
$$1 + \text{non-negative value} \ge 1 \ge 0$$
This case is trivial—no extra work needed here.

Case 2: $xy < 0$ #

Here, $xy$ is negative, so we need to show the negative term doesn't pull the entire expression below 0. Let's rewrite the inequality we need to prove:
$$1 \ge -\frac{xy}{e^{{\frac{1}{2}(x}2+y^2-2)}}$$
Since $xy < 0$, $-xy$ is positive. Multiply both sides by the positive exponential term (which preserves the inequality direction):
$$e^{{\frac{1}{2}(x}2+y^2-2)} \ge -xy$$

Now let $a = |x|$, $b = |y|$ (so $a,b > 0$). Then $-xy = ab$, and our inequality becomes:
$$e^{{\frac{1}{2}(a}2+b^2-2)} \ge ab$$

Step 1: Use the AM-GM Inequality #

From the Arithmetic Mean-Geometric Mean inequality, we know for non-negative $a,b$:
$$\frac{a^2 + b^2}{2} \ge ab$$
Let $t = \frac{a^2 + b^2}{2}$, so $t \ge ab$.

Step 2: Prove $e^{t-1} \ge t$ for all $t \ge 0$ #

Define the function $f(t) = e^{t-1} - t$. Let's find its minimum:

• Derivative: $f'(t) = e^{t-1} - 1$
• Set $f'(t) = 0$: $e^{t-1} = 1 \implies t = 1$
• Second derivative: $f''(t) = e^{t-1} > 0$ for all $t$, so $t=1$ is a global minimum.
• Evaluate $f(1) = e^0 - 1 = 0$, so $f(t) \ge 0$ for all $t \ge 0$, meaning $e^{t-1} \ge t$.

Step 3: Combine the Results #

Substitute $t = \frac{a^2 + b^2}{2}$ back into the inequality:
$$e^{{\frac{1}{2}(a}2+b^2)-1} = e^{t-1} \ge t \ge ab$$
This confirms $e^{{\frac{1}{2}(x}2+y^2-2)} \ge -xy$. Reversing our earlier steps shows $1 + \frac{xy}{e^{{\frac{1}{2}(x}2+y^2-2)}} \ge 0$ when $xy < 0$.

Conclusion #

Both cases confirm the inequality holds for all $x,y \in \mathbb{R^2}$. Your graph-based intuition was spot-on—this expression is always non-negative, and it converges to 1 as $x,y \to \pm\infty$ because the exponential term grows far faster than $xy$, making the fraction approach 0.

内容的提问来源于stack exchange，提问作者NewUser