Cauchy-Schwartz in disguise!

Question

Let $u\geqslant v\geqslant w\geqslant 0$ and $-1\leqslant a,b,c\leqslant 1$. Is the following inequality true? $$(u^2-2auv+v^2)(v^2-2bvw+w^2)\geqslant (v^2-auv-bvw+cuw)^2$$ $\textbf{What I have done this far:}$ We can rewrite $u^2-2auv+v^2=(v-au)^2+(1-a^2)u^2$, likewise we can write $v^2-2bvw+w^2=(v-bw)^2+(1-b^2)w^2$. Then it can be shown that the above inequality is equivalent to the following inequality $$((v-au)^2+(1-a^2)u^2)((v-bw)^2+(1-b^2)w^2)\geqslant ((v-au)(v-bw)+(c-ab)uw)^2$$ If $c\leqslant ab$ then it holds true since $a^2,b^2\leqslant 1$ imply that the left side not smaller than $(v-au)^2(v-bw)^2$. However when $c>ab$ it is no longer clear. For some special case like $c=1,a=b=0$ it is again true. My question is then, is the inequality true for any $c>ab$?

Answer 1

It doesn't hold.

In $$ (u^2-2auv+v^2)(v^2-2bvw+w^2)\geqslant (v^2-auv-bvw+cuw)^2 $$, let $u=v > 0$ and $a=1$ which gives $$ 0\geqslant (-bvw + cuw)^2 = (-b + c)^2 (uw)^2 $$ which is wrong for all $b \ne c$ and $w >0$.