Fie \( n\geq 3 \) un numar natural si \( a \) un numar real nenul. Aratati ca orice solutie nereala \( z \) a ecuatiei \( x^{n}+ax+1=0 \)satisface inegalitatea
\( |z| \geq\sqrt[n]{\frac{1}{n-1}} \).
Ecuatia x^n+ax+1 are solutie |z|>=1/\sqrt[n-1]{n-1}
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Ecuatia x^n+ax+1 are solutie |z|>=1/\sqrt[n-1]{n-1}
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
Din faptul ca \( x \) este solutie pentru ecuatia \( x^n+ax+1=0 \), atunci si \( \bar x \) este solutie deoarece ecuatia conjugata da \( \bar x^n+a\bar x+1=0 \).
Scadem cele doua ecuatii \( (x^n-\bar x^n)+a(x-\bar x)=0 \) si obtinem \( (x-\bar x)(\sum^{n-1}_{k=0} x^{n-1-k} \bar x^k +a)=0 \). Dar cum \( x\in C \) atunci \( x\not= \bar x \) si rezulta ca \( \sum^{n-1}_{k=0} x^{n-1-k} \bar x^k=-a \).
Scoatem pe \( a \) din ecuatie si obtinem \( \sum^{n-1}_{k=0} x^{n-1-k} \bar x^k=-\frac{1+x^n}{x} \) echivalent cu \( \sum^{n-1}_{k=1} x^{n-k} \bar x^k=-1 \).
Aplicand modulul si inegalitatea modulelor gasim: \( 1=|\sum^{n-1}_{k=1} x^{n-k} \bar x^k| \leq \sum^{n-1}_{k=1} |x^{n-k} \bar x^k|=(n-1)|x|^n \) de unde obtinem inegalitatea dorita \( |x| \geq \sqrt[n]{ \frac{1}{n-1}} \)
Scadem cele doua ecuatii \( (x^n-\bar x^n)+a(x-\bar x)=0 \) si obtinem \( (x-\bar x)(\sum^{n-1}_{k=0} x^{n-1-k} \bar x^k +a)=0 \). Dar cum \( x\in C \) atunci \( x\not= \bar x \) si rezulta ca \( \sum^{n-1}_{k=0} x^{n-1-k} \bar x^k=-a \).
Scoatem pe \( a \) din ecuatie si obtinem \( \sum^{n-1}_{k=0} x^{n-1-k} \bar x^k=-\frac{1+x^n}{x} \) echivalent cu \( \sum^{n-1}_{k=1} x^{n-k} \bar x^k=-1 \).
Aplicand modulul si inegalitatea modulelor gasim: \( 1=|\sum^{n-1}_{k=1} x^{n-k} \bar x^k| \leq \sum^{n-1}_{k=1} |x^{n-k} \bar x^k|=(n-1)|x|^n \) de unde obtinem inegalitatea dorita \( |x| \geq \sqrt[n]{ \frac{1}{n-1}} \)
Vrajitoarea Andrei