sailorlyts14eh

2022-09-30

Assume $n\ge 3,c\ge 1,d\ge 1$ are natural numbers such that $c²+d²-\left(n-1\right)cd<0$. Show that $\left(n³-n+1\right)c²+\left(n+1\right)d²-\left(n²+n-1\right)cd>1$

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typeOccutfg

Expert

Let $c,d$ be positive real numbers and $n>1$ (esp., ${n}^{3}-n+1>0$). By the arithmetic-geometric inequality
$\left({n}^{3}-n+1\right){c}^{2}+\left(n+1\right){d}^{2}\ge 2\cdot \sqrt{\left({n}^{3}-n+1\right)\left(n+1\right)}\cdot cd.$
One checks by multiplying out that
$4\left({n}^{3}-n+1\right)\left(n+1\right)=\left({n}^{2}+n-1{\right)}^{2}+3+3{n}^{2}\left({n}^{2}-1\right)+2n\left({n}^{2}+1\right),$
hence $2\cdot \sqrt{\left({n}^{3}-n+1\right)\left(n+1\right)}>{n}^{2}+n-1$ and finally
$\left({n}^{3}-n+1\right){c}^{2}+\left(n+1\right){d}^{2}>\left({n}^{2}+n-1\right)cd.$

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