Given positives $m, n, p$ such that $mnp = 1$, prove that $$\large \frac{m^2}{n + p^2} + \frac{n^2}{p + m^2} + \frac{p^2}{m + n^2} \ge \frac{m}{n^2 + p} + \frac{n}{p^2 + m} + \frac{p}{m^2 + n}$$
I can probably brutally force this inequality by the Buffalo Way, which I hate to use, but I won't.
Let $m = \dfrac{x}{y}, n = \dfrac{y}{z}, p = \dfrac{z}{x} (x, y, z > 0)$. The inequality becomes $$\frac{zx^4}{x^2y^3 + y^2z^3} + \frac{xy^4}{y^2z^3 + z^2x^3} + \frac{yz^4}{z^2x^3 + x^2y^3} \ge \frac{z^2x^2}{xy^3 + yz^3} + \frac{x^2y^2}{yz^3 + zx^3} + \frac{y^2z^2}{zx^3 + xy^3}$$
I could probably use the Chebyshev inequality several times, but the conditions needed contradict each other.
$$\frac{zx^4}{x^2y^3 + y^2z^3} + \frac{xy^4}{y^2z^3 + z^2x^3} + \frac{yz^4}{z^2x^3 + x^2y^3}$$
$$ \ge 2 \cdot \left[\frac{zx^4}{(x + y)(xy^3 + yz^3)} + \frac{xy^4}{(y + z)(yz^3 + zx^3)} + \frac{yz^4}{(z + x)(zx^3 + xy^3)}\right]$$
$$ \ge \frac{2 \cdot \left(\dfrac{z}{x + y} + \dfrac{x}{y + z} + \dfrac{y}{z + x}\right)\left(\dfrac{x^4}{xy^3 + yz^3} + \dfrac{y^4}{yz^3 + zx^3} + \dfrac{z^4}{zx^3 + xy^3}\right)}{3}$$
(It can be easily proven that $\dfrac{z}{x + y} + \dfrac{x}{y + z} + \dfrac{y}{z + x} \ge \dfrac{3}{2}$, using Titu's Lemma.)
$$\ge \frac{x^4}{xy^3 + yz^3} + \frac{y^4}{yz^3 + zx^3} + \frac{z^4}{zx^3 + xy^3}$$
For the final step to prove (unsuccessfully) this inequality, using the rearrangement inequality for $\left(\dfrac{x^2}{xy^3 + yz^3}, \dfrac{y^2}{yz^3 + zx^3}, \dfrac{z^2}{zx^3 + xy^3}\right)$ and $(x^2, y^2, z^2)$, we have that $$\frac{x^4}{xy^3 + yz^3} + \frac{y^4}{yz^3 + zx^3} + \frac{z^4}{zx^3 + xy^3} \ge \frac{x^2y^2}{xy^3 + yz^3} + \frac{y^2z^2}{yz^3 + zx^3} + \frac{z^2x^2}{zx^3 + xy^3}$$
By the way, the conditions for the first, second and fourth are as followed:
$x + y \ge y + z \ge z + x$ and $xy^3 + yz^3 \le yz^3 + zx^3 \le zx^3 + xy^3$
$\dfrac{z}{x + y} \le \dfrac{x}{y + z} \le \dfrac{y}{z + x}$ and $\dfrac{x^4}{xy^3 + yz^3} \le \dfrac{y^4}{yz^3 + zx^3} \le \dfrac{z^4}{zx^3 + xy^3}$
$\dfrac{x^2}{xy^3 + yz^3} \le \dfrac{y^2}{yz^3 + zx^3} \le \dfrac{z^2}{zx^3 + xy^3}$ and $x^2 \le y^2 \le z^2$