Problem 1354

To show that for any real number $x$:
$$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0,$$
we will analyze the expression step by step.

1. Rewrite the expression:
$$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} = x^2 (\sin{x} + 1) + x \cos{x} + \frac{1}{2}.$$

2. **Analyze the coefficient of $x^2$:**
The coefficient of $x^2$ is $\sin{x} + 1$. Since $\sin{x}$ ranges from $-1$ to $1$, we have:
$$-1 + 1 \leq \sin{x} + 1 \leq 1 + 1,$$
which simplifies to:
$$0 \leq \sin{x} + 1 \leq 2.$$
Therefore, $\sin{x} + 1 \geq 0$.

3. Consider the discriminant of the quadratic expression:
The quadratic expression in $x$ is:
$$x^2 (\sin{x} + 1) + x \cos{x} + \frac{1}{2}.$$
The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c$ is given by:
$$\Delta = b^2 - 4ac.$$
Here, $a = \sin{x} + 1$, $b = \cos{x}$, and $c = \frac{1}{2}$. Thus,
$$\Delta = (\cos{x})^2 - 4(\sin{x} + 1)\left(\frac{1}{2}\right).$$
Simplifying,
$$\Delta = \cos^2{x} - 2(\sin{x} + 1).$$
$$\Delta = \cos^2{x} - 2\sin{x} - 2.$$

4. Show that the discriminant is always negative:
We know that $\cos^2{x} = 1 - \sin^2{x}$. Substituting this in,
$$\Delta = 1 - \sin^2{x} - 2\sin{x} - 2.$$
$$\Delta = -\sin^2{x} - 2\sin{x} - 1.$$
Since $\sin^2{x} \geq 0$, $-\sin^2{x} \leq 0$. Therefore,
$$\Delta = -(\sin^2{x} + 2\sin{x} + 1).$$
$$\Delta = -(\sin{x} + 1)^2.$$
Since $(\sin{x} + 1)^2 \geq 0$, we have:
$$\Delta \leq 0.$$

5. Conclusion from the discriminant:
Since the discriminant $\Delta \leq 0$, the quadratic expression $x^2 (\sin{x} + 1) + x \cos{x} + \frac{1}{2}$ does not have real roots. This implies that the quadratic expression does not cross the x-axis and is either always positive or always negative.

6. **Evaluate the expression at $x = 0$:**
$$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} \bigg|_{x=0} = 0^2 \sin{0} + 0 \cos{0} + 0^2 + \frac{1}{2} = \frac{1}{2}.$$
Since the expression is positive at $x = 0$ and the quadratic expression does not cross the x-axis, it must be positive for all $x$.

The final answer is $\boxed{ x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0 }$.