Inequality with a precondition

Question

Show that if $a^2+b^2+c^2=1$ with $a, b, c\gt0$, then $$S=\frac{1-(a^4+b^4+c^4)}{abc} \ge 2*\sqrt 3$$ I noticed that the expression on the left hand side is equivalent to $2*(ab/c+ac/b+cb/a)$. By applying AM-GM to each two of the terms and adding the results, I got $S\ge 2*(a+b+c)$. Applying the CS inequality yielded $S\ge 2abc$. However, both results are too weak. I also managed to show the required result by expressing c in terms of a and b and then differentiating twice. However, this took me pages and was very unaesthetic.

Answer 1

We need to prove that $$(a^2+b^2+c^2)^2-a^4-b^4-c^4\geq2abc\sqrt{3(a^2+b^2+c^2)}$$ or $$a^2b^2+a^2c^2+b^2c^2\geq abc\sqrt{3(a^2+b^2+c^2)}$$ or $$a^4(b^2-c^2)^2+b^4(a^2-c^2)^2+c^4(a^2-b^2)^2\geq0.$$