Prove the inequality $(x+y+z)^2\ge 4(x^2y^2+y^2z^2+z^2x^2)$

Question

In $\Delta ABC$,let $x=\sin{A},y=\sin{B},z=\sin{C}$,show that $$(x+y+z)^2\ge 4(x^2y^2+y^2z^2+z^2x^2)$$

I tried C-S and more, but without success.

I am looking for an human proof, which we can use during competition.

Answer 1

We need to prove that $$\left(\sum\limits_{cyc}\frac{2S}{ab}\right)^2\geq4\sum\limits_{cyc}\frac{16S^4}{a^4b^2c^2}$$ or $$a^2b^2c^2(a+b+c)^2\geq(a^2b^2+a^2c^2+b^2c^2)\sum\limits_{cyc}(2a^2b^2-a^4)$$ or $$\sum\limits_{cyc}(a^6b^2+a^6c^2-2a^4b^4-2a^4b^2c^2+2a^3b^3c^2)\geq0$$ or $$\sum\limits_{cyc}(a-b)^2(a^2b^2(a+b)^2-a^2b^2c^2)\geq0$$ or $$\sum\limits_{cyc}(a-b)^2a^2b^2(a+b+c)(a+b-c)\geq0$$ and we are done!