nierówność

[anilewe_MM]

Czy z \(x^6-\frac{1}{x^6}>1\) dla dla x różnego od zera wynika \(x^{18}+\frac{1}{x^{18}}>2x^6+\frac{2}{x^6}\) ?

[Icanseepeace]

\( L = x^{18} + \frac{1}{x^{18}} = (x^6 + \frac{1}{x^6})(x^{12} - 1 + \frac{1}{x^{12}}) = (x^6 + \frac{1}{x^6})((x^6 - \frac{1}{x^6})^2 -1) + 2) > 2(x^6 + \frac{1}{x^6}) = P \)

[Jerry]

Albo, ze "szkolnym" zapisem:
\[x^6-\frac{1}{x^6}>1\qquad|^2\\ \Downarrow\\
x^{12}-2+\frac{1}{x^{12}}>1\\
x^{12}-1+\frac{1}{x^{12}}>2\qquad|\cdot\left(x^6+\frac{1}{x^6}\right)\\
x^{18}+\frac{1}{x^{18}}>2x^6+\frac{2}{x^6}\\ \text{CKD}\]
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