- Secondary
- Calculus and Analysis
- Precalculus
Polynomials

sailorlyts14eh

Answered

2022-09-30

Assume $n\ge 3,c\ge 1,d\ge 1$ are natural numbers such that $c\xb2+d\xb2-(n-1)cd<0$. Show that $(n\xb3-n+1)c\xb2+(n+1)d\xb2-(n\xb2+n-1)cd>1$

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Recalculate according to your conditions!

Answer & Explanation

typeOccutfg

Expert

2022-10-01Added 6 answers

Let $c,d$ be positive real numbers and $n>1$ (esp., ${n}^{3}-n+1>0$). By the arithmetic-geometric inequality

$({n}^{3}-n+1){c}^{2}+(n+1){d}^{2}\ge 2\cdot \sqrt{({n}^{3}-n+1)(n+1)}\cdot cd.$

One checks by multiplying out that

$4({n}^{3}-n+1)(n+1)=({n}^{2}+n-1{)}^{2}+3+3{n}^{2}({n}^{2}-1)+2n({n}^{2}+1),$

hence $2\cdot \sqrt{({n}^{3}-n+1)(n+1)}>{n}^{2}+n-1$ and finally

$({n}^{3}-n+1){c}^{2}+(n+1){d}^{2}>({n}^{2}+n-1)cd.$

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