# A parametric polynomial inequality

Recently I had use for the following inequality, which was proven in a rather inelegant way:

For all $y \in [0,1]$ and all integers $0\leq i \leq s$, the following inequality holds: \begin{align*}\left(2+2y^i-4y^s+y^{2s-i}\right)^s \leq \left(2-y^s\right)^{2s-i}. \end{align*}

I now wonder:
1. Is this a special case of some known inequality?
2. Does anyone know some nice proof for this and similar inequalities in one or more variables?

Notice that at the endpoint cases $i=0, i=s$ we have equality. Therefore to prove the inequality it is enough to show that the map $$i \mapsto \ln(2+2y^{i}-4y^{s}+y^{2s-i}), \quad 0\leq i \leq s$$ is convex. We have $$\frac{d^{2}}{di^{2}} \ln(2+2y^{i}-4y^{s}+y^{2s-i})=\frac{2y^{2i}(\ln(y))^{2}(4y^{2(s-i)}-4y^{s-i}-2y^{3(s-i)}+y^{-i}(2+y^{2(s-i)}))}{(2+2y^{i}-4y^{s}+2y^{2s-i})^{2}}$$ If you denote $y^{s-i}=a$ where $0\leq a\leq 1$ then $$4y^{2(s-i)}-4y^{s-i}-2y^{3(s-i)}+y^{-i}(2+y^{2(s-i)}) \geq 4a^{2}-4a-2a^{3}+2+a^{2}$$ and the minimal value of the latter expression when $a \in [0,1]$ is $\frac{26}{27}> 0$
• I'm confused. The LHS is convex in $i$, but so is the the RHS... So couldn't the RHS be more convex and thus the inequality fail? – Yaakov Baruch Aug 28 '17 at 20:04
• I got it. after taking $\text{ln}()$ the RHS is linear... Nice proof! – Yaakov Baruch Aug 28 '17 at 20:16