Problem 1783

Proof: A cursory look at this inequality might make you hesitate. The relationship between variables $a, b, c, d$ is difficult to manipulate; moreover, the inequality contains square roots, which adds to the complexity. Surprisingly, a simple method is to use Chebyshev's inequality to unravel this mystery. Let's explore this method.
By assumption, we have $\square$
$$\sum_{c y c} \frac{1}{a}=\sum_{c y c} a \Leftrightarrow \sum_{c y c}\left(a-\frac{1}{a}\right)=0 \Leftrightarrow \sum_{c y c}\left(\frac{a^{2}-1}{a}\right)=0$$

The inequality is transformed as follows:
$$\sum_{c y c}\left(2 a-\sqrt{a^{2}+3}\right) \geq 0 \Leftrightarrow \sum_{c y c} \frac{a^{2}-1}{2 a+\sqrt{a^{2}+3}} \geq 0$$

How to proceed? The idea is to apply Chebyshev's inequality to the following sequences:
$$\left(a^{2}-1, b^{2}-1, c^{2}-1, d^{2}-1\right) ;\left(\frac{1}{2 a+\sqrt{a^{2}+3}}, \frac{1}{2 b+\sqrt{b^{2}+3}}, \frac{1}{2 c+\sqrt{c^{2}+3}}, \frac{1}{2 d+\sqrt{d^{2}+3}}\right)$$

However, this idea fails because the first sequence is increasing, while the second sequence is decreasing. Using Chebyshev's inequality would reverse the direction of the inequality. We hope it's not the case. Noting that $\sum_{c y c}\left(\frac{a^{2}-1}{a}\right)=0$, we will change the inequality to the following form:
$$\sum_{c y c} \frac{a^{2}-1}{a} \cdot \frac{a}{2 a+\sqrt{a^{2}+3}} \geq 0$$

Assuming $a \geq b \geq c \geq d$, using the identity $\frac{a}{2 a+\sqrt{a^{2}+3}}=\frac{1}{2+\sqrt{1+\frac{3}{a^{2}}}}$, we examine:
$$\left(\frac{a^{2}-1}{a}, \frac{b^{2}-1}{b}, \frac{c^{2}-1}{c}, \frac{d^{2}-1}{d}\right)$$

and
$$\left(\frac{a}{2 a+\sqrt{a^{2}+3}}, \frac{b}{2 b+\sqrt{b^{2}+3}}, \frac{c}{2 c+\sqrt{c^{2}+3}}, \frac{d}{2 d+\sqrt{d^{2}+3}}\right)$$

These are two increasing sequences. Thus, by Chebyshev's inequality, we get:
$$\sum_{c y c}\left(\frac{a^{2}-1}{a}\right) \cdot\left(\frac{a}{2 a+\sqrt{a^{2}+3}}\right) \geq \frac{1}{4}\left(\sum_{\text {cyc }} \frac{a^{2}-1}{a}\right)\left(\sum_{\text {cyc }} \frac{a}{2 a+\sqrt{a^{2}+3}}\right)=0$$

Proof complete, equality holds when $a=b=c=d=1$.
The key to this simple proof lies in appropriately decomposing the numerator and denominator of the fraction using the given conditions and the order that satisfies Chebyshev's inequality. Based on this method, we construct the following general approach:

Assume we need to prove the inequality (expressed as a sum of fractions):
$$\frac{x_{1}}{y_{1}}+\frac{x_{2}}{y_{2}}+\cdots+\frac{x_{n}}{y_{n}} \geq 0$$

where $x_{1}, x_{2}, \cdots, x_{n}$ are real numbers, and $y_{1}, y_{2}, \cdots, y_{n}$ are positive real numbers.
Generally, any inequality can be transformed into this form. If some fractions have a negative denominator, we multiply both the numerator and the denominator by -1 to obtain a new sequence of positive numbers $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ such that the sequence
$$\left(a_{1} x_{1}, a_{2} x_{2}, \cdots, a_{n} x_{n}\right)$$

is increasing, but the sequence
$$\left(a_{1} y_{1}, a_{2} y_{2}, \cdots, a_{n} y_{n}\right)$$

is decreasing. Applying Chebyshev's inequality, we have:
$$\sum_{i=1}^{n} \frac{x_{i}}{y_{i}} \geq \frac{1}{n}\left(\sum_{c y c} a_{i} x_{i}\right) \sum_{c y c}\left(\frac{1}{a_{i} y_{i}}\right)$$
$$\sum_{i=1}^{n} a_{i} x_{i} \geq 0$$