Minimum is $3$.

Substitute $a^2 = \frac{x^2}{x^2+y^2+z^2}, b^2 = \frac{y^2}{x^2+y^2+z^2}, c^2 = \frac{z^2}{x^2+y^2+z^2}$. Then $S = (x^2+y^2+z^2)(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2})-2\frac{x^3+y^3+z^3}{xyz} = 3+\sum_{cyc} x^2(\frac{1}{y}-\frac{1}{z})^2\geq3$ Since $S = 3$ at $a = b = c = \frac{\sqrt{3}}{3}$, 3 is the minimum of $S$.

Let $a, b, c$ be positive reals so that $a^2+b^2+c^2=1$. Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{2(a^3+b^3+c^3)}{abc}\geq 3$$

https://artofproblemsolving.com/community/q1h1872215p12705389 similar but for min

sqing wrote: Let $a, b, c$ be positive reals so that $a^2+b^2+c^2=1$. Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{2(a^3+b^3+c^3)}{abc}\geq 3$$ Let $a, b, c$ be positive reals so that $a+b+c=1.$ Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} - \frac{2(a^3+b^3+c^3)}{abc}\leq 3$$

Let $a, b, c$ be positive real numbers with $a^2+b^2+c^2=12.$ Prove that$$\frac{a^4}{\sqrt{a^3+1}}+\frac{b^4}{\sqrt{b^3+1}}+\frac{c^4}{\sqrt{c^3+1}}\geq 16.$$Let $a, b, c$ be positive real numbers with $a+b+c=6.$ Prove that$$\frac{a^2}{\sqrt{a^3+1}}+\frac{b^2}{\sqrt{b^3+1}}+\frac{c^2}{\sqrt{c^3+1}}\geq \frac{144}{a^3+b^3+c^3+12}.$$Let $a, b, c$ be positive real numbers with $a^2+b^2+c^2\leq 12.$ Prove that$$(a^3+1)(b^3+1)(c^3+1)\leq {27}^2.$$(George Apostolopoulos)

Let $x, y, z $ be positive real numbers such that $x^4+y^4+c^4=3.$ Prove that$$\sqrt{\frac{yz}{7-2x}}+\sqrt{\frac{zx}{7-2y}}+\sqrt{\frac{xy}{7-2z}}\leq \frac{3}{\sqrt 5}.$$(Hoang Le Nhat Tung)

sqing wrote: Let $x, y, z $ be positive real numbers such that $x^4+y^4+z^4=3.$ Prove that$$\sqrt{\frac{yz}{7-2x}}+\sqrt{\frac{zx}{7-2y}}+\sqrt{\frac{xy}{7-2z}}\leq \frac{3}{\sqrt 5}.$$(Hoang Le Nhat Tung) AM-GM and Jensen.

sqing wrote: Let $x, y, z $ be positive real numbers such that $x^4+y^4+c^4=3.$ Prove that$$\sqrt{\frac{yz}{7-2x}}+\sqrt{\frac{zx}{7-2y}}+\sqrt{\frac{xy}{7-2z}}\leq \frac{3}{\sqrt 5}.$$(Hoang Le Nhat Tung)

sqing wrote: Let $x, y, z $ be positive real numbers such that $x^4+y^4+c^4=3.$ Prove that$$\sqrt{\frac{yz}{7-2x}}+\sqrt{\frac{zx}{7-2y}}+\sqrt{\frac{xy}{7-2z}}\leq \frac{3}{\sqrt 5}.$$(Hoang Le Nhat Tung) nice solution sqing here's another approach. consider the function $f(x)=\sqrt{x}$ the second derivative is $\frac{-1}{4x^{\frac32}}$ and it is trivial that it is always negetive so by $\textit{Jensen's inequality}$ it remains to prove that: \[\frac35 \ge \sum_{cyc} \frac{xy}{7-2z} \] Using $AM-GM$ and $\textit{CS}$ we have: \begin{align*} \sum_{cyc} \frac{xy}{7-2z} \le& \sum_{cyc} \frac{5-z^4}{22-2z^4}\\ =& \sum_{cyc} \frac12- \frac{3}{11-z^4}\\ =& \frac32 +\sum_{cyc} \frac{3}{11-z^4}\\ \le& \frac32 - 3(\frac{9}{33-\sum_{cyc} x^4})=\frac35 \quad \blacksquare \end{align*}as desired