Problem 1732

Prove that the original inequality is equivalent to
$$\frac{a^{4}+b^{4}+c^{4}+d^{4}}{a b c d} \geqslant \frac{(2 k-a)^{4}+(2 k-b)^{4}+(2 k-c)^{4}+(2 k-d)^{4}}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)}$$

is equivalent to $\frac{\left(a^{2}-b^{2}\right)^{2}+\left(c^{2}-d^{2}\right)^{2}+2\left(a^{2} b^{2}+c^{2} d^{2}\right)}{a b c d} \geqslant$
$$\frac{\left[(2 k-a)^{2}-(2 k-b)^{2}\right]^{2}+\left[(2 k-c)^{2}-(2 k-d)^{2}\right]^{2}+2\left[(2 k-a)^{2}(2 k-b)^{2}+(2 k-c)^{2}(2 k-d)^{2}\right]}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)} .$$

Assume without loss of generality that $a \geqslant b \geqslant c \geqslant d$, and we will prove the following three inequalities:
$$\begin{array}{c} \frac{\left(a^{2}-b^{2}\right)^{2}}{a b c d} \geqslant \frac{\left[(2 k-a)^{2}-(2 k-b)^{2}\right]^{2}}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)}, \\ \frac{\left(c^{2}-d^{2}\right)^{2}}{a b c d} \geqslant \frac{\left[(2 k-c)^{2}(2 k-d)^{2}\right]^{2}}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)} \\ \frac{2\left(a^{2} b^{2}+c^{2} d^{2}\right)}{a b c d} \geqslant \frac{2\left[(2 k-a)^{2}(2 k-b)^{2}+(2 k-c)^{2}(2 k-d)^{2}\right]}{(2 k-a)(2 k-b)(2 k-c)(2 k-d)} \end{array}$$

Since $a, b, c, d \leqslant k$, we have
$$\begin{array}{l} (2 k-a)(2 k-b) \geqslant a b, \\ (2 k-c)(2 k-d) \geqslant c d, \end{array}$$

Thus, it suffices to prove $\quad \frac{\left(a^{2}-b^{2}\right)^{2}}{a b} \geqslant \frac{\left[(2 k-a)^{2}-(2 k-b)^{2}\right]^{2}}{(2 k-a)(2 k-b)}$,
hence
$$\begin{aligned} \frac{\left(a^{2}-b^{2}\right)^{2}}{a b} & \geqslant \frac{\left[(2 k-a)^{2}-(2 k-b)^{2}\right]^{2}}{(2 k-a)(2 k-b)}(a-b)^{2}\left[(a+b)^{2}(2 k-a)(2 k-b)\right] \\ & \geqslant(a-b)^{2}[(4 k-a-b) 2 a b]\left(\frac{a+b}{2}\right)^{2}(2 k-a)(2 k-b) \\ & \geqslant\left(2 k-\frac{a+b}{2}\right)^{2} a b\left(\frac{a+b}{2}\right)^{2}\left[\left(2 k-\frac{a+b}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}\right] \\ \geqslant & \left(2 k-\frac{a+b}{2}\right)^{2}\left[\left(\frac{a+b}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}\right]\left[\left(2 k-\frac{a+b}{2}\right)^{2}-\right. \\ & \left.\left(\frac{a+b}{2}\right)^{2}\right]\left(\frac{a-b}{2}\right)^{2} \geqslant 0 \end{aligned}$$

By $k \geqslant a \geqslant b$, we know $2 k-\frac{a+b}{2} \geqslant \frac{a+b}{2}$, so (5) is clearly true, hence (2) is true. Similarly, (3) can be proven.

By $k \geqslant a \geqslant b \geqslant c \geqslant d>0$, we know
$$\frac{a b}{c d} \geqslant 1, \frac{(2 k-c)(2 k-d)}{(2 k-a)(2 k-b)} \frac{a b}{c d} \geqslant 1$$

Let $f(x)=x+\frac{1}{x}$, then $f(x)=f\left(\frac{1}{x}\right)$, and $f(x)$ is monotonically increasing on $[1,+\infty)$, so
$$\text { (4) } \begin{aligned} \text { is equivalent to } & f\left(\frac{a b}{c d}\right) \geqslant f\left[\frac{(2 k-c)(2 k-d)}{(2 k-a)(2 k-b)}\right] \\ & \Leftrightarrow \frac{a b}{c d} \geqslant \frac{(2 k-c)(2 k-d)}{(2 k-a)(2 k-b)} \\ & \Leftrightarrow \frac{a b(2 k-a)(2 k-b)}{c d(2 k-c)(2 k-d)} \geqslant 1 \\ & \Leftrightarrow \frac{k^{2}-(k-a)^{2}}{k^{2}(k-c)^{2}} \cdot \frac{k^{2}-(k-b)^{2}}{k^{2}(k-d)^{2}} \geqslant 1 . \end{aligned}$$

By $k \geqslant a \geqslant b \geqslant c \geqslant d>0$, we have
$$k \geqslant k-c \geqslant k-a>0, k \geqslant k-d \geqslant k-b>0$$
(6) is true, hence (4) is true.

Adding (2), (3), and (4) yields (1), thus the original inequality is proven.