Question

Daca x este un numar real din intervalul [0;1], demonstrati ca $\sqrt{(x-1 )^{2} } + x^{100} +x \geq 1$

Answer 1

   
[tex]\displaystyle\\ \sqrt{(x-1 )^{2} } + x^{100} +x \geq 1 \\\\ x\in [0;~1] ~~~\Longrightarrow~~~ \boxed{0 \leq x^{100} \leq 1}\\\\ \sqrt{(x-1)^2} = \Big|x-1\Big| = \boxed{1-x} ~~~\text{deoarece: }~~ x \leq 1\\\\ \text{Rezolvare:}\\\\ \sqrt{(x-1 )^{2} } + x^{100} +x = \\\\ = \Big|x-1\Big| + x^{100} + x = \\\\ = 1-x + x^{100} +x = \\\\ =1\underline{-x}+\underline{x} +x^{100} = \boxed{\bf \boxed{\bf 1 + x^{100} \geq 1}}[/tex]