Let $\sigma_1\le\sigma_2,\sigma_1^*<\sigma_2^*$ be positive real numbers and suppose that $\sigma_1 \sigma_2=\sigma_1^* \sigma_2^*$, and that $\sigma_1^*+\sigma_2^*=1$.
Is it true that $$ 1-(\sigma_1+\sigma_2) \ge (\sigma_1^*)^2+(\sigma_2^*)^2-\sigma_1^*\sigma_1-\sigma_2^*\sigma_2 \,\,\,\,?$$
A more symmetric equivalent rewriting is
$$ 1-(\sigma_1+\sigma_2) \ge \sigma_1^*(\sigma_1^*-\sigma_1)+\sigma_2^*(\sigma_2^*-\sigma_2).$$
From this form, one can easily deduce that equality holds in the extreme case where $\sigma_1+\sigma_2=1=\sigma_1^*+\sigma_2^*$; indeed, in that case the LHS is zero. Since the sum and the product of $\sigma_1,\sigma_2$ $\sigma_1^*,\sigma_2^*$ are now identical, it follows that they are the same up to a permutation $1 \iff 2$. Since we have fixed their order in advance, we must have $\sigma_i=\sigma_i^*$, so the RHS also vanishes.
It's wrong. Try $\sigma_1=\sigma_2=\frac{1}{3},$ $\sigma_1^*=\frac{3-\sqrt5}{6}$ and $\sigma_2^*=\frac{3+\sqrt5}{6}$
But for $\sigma_1+\sigma_2\geq4\sigma_1\sigma_2$ the reversed inequality is true.
Indeed, let $\sigma_1=a$, $\sigma_2=b$, where $b\geq a>0$ and $a+b\geq4ab$.
Thus, $\sigma_1^*$ and $\sigma_2^*$ are roots of the equation: $$x^2-x+ab=0,$$ which gives $$\sigma_1^*=\frac{1-\sqrt{1-4ab}}{2}$$ and $$\sigma_2^*=\frac{1+\sqrt{1-4ab}}{2}.$$ Id est, we need to prove that: $$1-a-b+a\cdot\frac{1-\sqrt{1-4ab}}{2}+b\cdot\frac{1+\sqrt{1-4ab}}{2}\leq1-2ab$$ or $$(b-a)\sqrt{1-4ab}\leq a+b-4ab$$ or $$(a-b)^2(1-4ab)\leq(a+b-4ab)^2$$ or $$ab(a+b-1)^2\geq0$$ and we are done!