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\begin{document}

\begin{thm}
$$ \forall n\in \mathbb{N} \forall x>-1 : (1+x)^n\geq 1+nx $$
\end{thm}
\begin{proof}
באינדוקציה: עבור $n=1 $ נקבל ש- $1+x\geq 1+x$ שזה כמובן נכון. נניח שהטענה נכונה עבור $n$ כללי ונראה שעבור $n+1$ מתקיים
$$(1+x)^{n+1}=(1+x)^n\cdot (1+x) \geq (1+nx)(1+x)=1+nx+x+nx^2=$$
$$1+(n+1)x+nx^2\geq 1+(n+1)x $$
\end{proof}
\end{document}
