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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?

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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 $\frac{26}{27}> 0$

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  • $\begingroup$ 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? $\endgroup$ – Yaakov Baruch Aug 28 '17 at 20:04
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    $\begingroup$ I got it. after taking $\text{ln}()$ the RHS is linear... Nice proof! $\endgroup$ – Yaakov Baruch Aug 28 '17 at 20:16

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