Show that 1 + x &lt; 1 + x 2 if x &gt; 0

Why not just square both sides? 1 + x &lt; 1 + x + x 2 4 is trivial.

We have that for all x &gt; 0, we have 1 + x &lt; 1 + x + x 2 4 = ( 1 + x 2 ) 2 ,and hence 1 + x &lt; 1 + x 2 .

Why not just square both sides? 1 + x &lt; 1 + x + x 2 4 is trivial.

We have that for all x &gt; 0, we have 1 + x &lt; 1 + x + x 2 4 = ( 1 + x 2 ) 2 ,and hence 1 + x &lt; 1 + x 2 .

Check the solution to "Show that 1 + x < 1 +..."

Nathanial Levine

2022-09-29

Show that

\sqrt{1 + x} < 1 + \frac{x}{2}

x > 0

Jordan Owen

Beginner2022-09-30Added 7 answers

Why not just square both sides?

1 + x < 1 + x + \frac{x^{2}}{4}

is trivial.

gsragator9

Beginner2022-10-01Added 2 answers

We have that for all

x > 0

, we have

1 + x < 1 + x + \frac{x^{2}}{4} = {(1 + \frac{x}{2})}^{2},

and hence

\sqrt{1 + x} < 1 + \frac{x}{2} .

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