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Let $a\geq 0, x\geq 0, c\geq 0, y \geq 0$, and $0 \leq \gamma \leq 1/4$ be real numbers such that $$ a+x+c+y=1,$$ $$(a+c)^2+(x+y)^2+2cx \leq 1-2\gamma,$$ $$ \log(4) c+\log(3)a+\log(2)x \geq \frac{\log(6)}{2}+\frac{\sqrt{1-4\gamma}}{2}\log(3/2),$$ and $$a+c \geq 4/5.$$

I want to show that $a-x\leq 3(y-c)+\sqrt{1-4\gamma}$. Here is what I have so far:

The first two constraints give us $(a+c)^2+(1-a-c)^2\leq 1-2\gamma$, or equivalently, $(x+y)^2+(1-x-y)^2\leq 1-2\gamma$ which implies that $$ \frac{1-\sqrt{1-4\gamma}}{2}-c \leq a \leq \frac{1+\sqrt{1-4\gamma}}{2}-c \quad \quad \quad (1)$$ and $$ \frac{1-\sqrt{1-4\gamma}}{2}-y \leq x \leq \frac{1+\sqrt{1-4\gamma}}{2}-y. \quad \quad \quad (2)$$

By subtracting (2) from (1) we obtain $a-x \leq y-c+\sqrt{1-4\gamma}$, which is close to what we want. We'd be done if we knew that $y-c+\sqrt{1-4\gamma} \leq 3(y-c)+\sqrt{1-4\gamma}$, but this is only true if $y \geq c$. I haven't been able to show that, though.

I haven't used the 3rd and 4th constraints at the top. I've been trying to use Mathematica and Maple for this, but Mathematica can't handle all the constraints, and Maple gives me a bunch of crazy answers. Some people are very good at computing. I would immensely appreciate your help. Thank you.

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This will be too long to fit as a comment, so I'll post it here as a partial answer. It seems in your approach you have omitted the $2cx$ term from the second constraint when obtaining your inequalities $(1)$ and $(2)$. They are still valid since $cx \geq 0$, though now they may not be tight enough to give the result.

Note that taking the third constraint and rewriting it in terms of $\log{3}$ and $\log{2}$, we obtain \begin{equation} (4c + 2x - 1 + \sqrt{1-4\gamma})\log{2} \geq (1 + \sqrt{1-4\gamma} - 2a)\log{3}, \tag*{(1)}\end{equation} and since $\dfrac{\log{2}}{\log{3}} < 1$, this implies that \begin{equation*} 1 - 2a < 4c + 2x -1. \end{equation*} Using the first constraint, this becomes $y < c$, so your suggested approach (as it stands) won't work out.

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  • $\begingroup$ Thank you. Do we necessarily have the strict inequality $1-2a<4c+2x-1$? Perhaps we have $y \leq c$. If I can show that $y=c$ that works, too. $\endgroup$ – Steve Mar 25 at 23:52
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    $\begingroup$ Yes, I think the inequality will be strict since it comes from the strict inequality $\log{2}/\log{3} < 1$ applied to my inequality $(1)$ after dividing by $\log{3}$ $\endgroup$ – Zac Mar 25 at 23:59

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