Let $x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n})$. Prove that $x_{n+1} < x_{n}$

Hint: $$x_n- x_{n+1}=x_n-\frac{1}{2}\left(x_n+\frac{a}{x_n}\right)=\frac{1}{2} \frac{x_n^2-a}{x_n} \geq 0$$ since $x_n^2 \geq a$ (?)

Lemma: $x_n>\sqrt{a}$ for all $n\ge 1$.

Proof: The proof is by induction. Suppose $x_n>\sqrt{a}$. Divding $a$ by both sides, you get $\frac{a}{x_n}<\sqrt{a}$. Combining these, you get $x_n>\frac a{x_n}$, so $\sqrt{x_n}>\sqrt{\frac a{x_n}}$, so $$ \left(\sqrt{x_n}-\sqrt{\frac a{x_n}}\right)^2>0 $$ This rearranges to $$ \frac12\left(x_n+\frac{a}{x_n}\right)>\sqrt{a} $$ which is exactly $x_{n+1}>\sqrt{a}$. $\hspace{.2cm}\square$

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Now we know $x_n>\sqrt{a}$ for all $n$, which as before implies $x_n>\frac{a}{x_n}$. Conclude with $$ x_{n}=\frac12 x_n+\frac12 x_n> \frac12 x_n+\frac12\cdot \frac{a}{x_{n}}=x_{n+1} $$

I suppose $a>0$. Now consider the function $f(x)=\frac12\left(x+\frac{a}{x}\right)$ for $x>\sqrt{a}$. Take the derivative and show that the derivative is positive then deduce that the function is an increasing one. After that start an induction. If $x_n>x_{n+1}$ then $f(x_n)>f(...$