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I am a little stuck on this question and would appreciate some help. The question asks me to prove that $\sqrt{1+x}<1+\frac{x}{2}$ for all $x>0$.

I squared both sides of the question to get $1+x<\frac{x^2}{4}+x+1$ for all $x>0$. Then, I multiplied both sides by $4$ to get $4+4x<x^2+4x+4$ for all $x>0$.

I am a little stuck and was wondering what to do after this step and how to actually provide sufficient proof to say that this statement is true.

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You did well. Let's finish it: we have $x^2>0$ thus $\dfrac{x^2}{4}>0$. Adding $x+1$ to both sides, we have $$\dfrac{x^2}{4}+x+1> x+1$$ or $$(\frac{x}{2}+1)^2>x+1$$ or $$\frac{x}{2}+1>\sqrt{x+1}$$

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Squaring the equation was sufficient. After you cancel the $1+x$ on each side, you have

$$\frac{x^2}{4} > 0$$

which is true for all real $x \ne 0$ since $x^2 \ge 0$ (with equality only when $x=0$). Thus, the inequality is proved.

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You can also prove the inequality using the Mean Value Theorem. Define $f(t)=\sqrt{t+1}$ and apply the theorem on the interval $[0,x]$. Then, there exists $c\in(0,x)$ s.t. $$\frac{\sqrt{x+1}-\sqrt{1}}{x-0}=\frac{1}{2\sqrt{c+1}}>\frac{1}{2}.$$

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I am answering less to provide a list of steps to follow for proving the inequality, and more to discuss how I believe that the argument should be presented. This seems appropriate, as the question has been tagged with and . If I asked a student to prove some inequality and their first step was to assume the inequality and square both sides, I would almost certainly deduct points unless the student were careful about justifying such a step. The goal here is to start with known true statements and to then deduce the desired result.

Claim: For any $x > 0$, $$ \sqrt{1+x} < 1 + \frac{x}{2}. $$

Proof: Assume that $x > 0$. The square of any real number is positive, hence $$ 0 < \left( \frac{x}{2} \right)^2. $$ By additive cancelation (i.e. by "adding the same number to both sides"), this implies that $$ x + 1 < \left( \frac{x}{2} \right)^2 + x + 1 = \left( \frac{x}{2} + 1 \right)^2. $$ The square root function is increasing on $[0,\infty)$ (that is, if $0 \le a < b$, then $0 \le \sqrt{a} \le \sqrt{b}$), and the assumption that $x > 0$ ensures that $1+x > 0$, thus $1 + x$ is in the domain of the square root function (i.e. it has a well-defined real square root). Hence $$ \sqrt{x+1} < \sqrt{ \left( 1 + \frac{x}{2} \right)^2 } = \left|1 + \frac{x}{2}\right| = 1 + \frac{x}{2}, $$ where the final equality follows from the assumption that $x > 0$.

Discussion

Remember that if you start with false assumptions, you can prove anything—false implies true. Thus it is poor practice to begin a proof by asserting the statement which you are trying to prove and then applying algebraic manipulation. You can do this if you are careful—be very observant of two-sided implications and check the hypotheses at each step. Indeed, this is a very reasonable way of deducing a correct proof in the first place.

However, when presenting an argument, such a proof is typically poor style. Personally, I think that it is better (stylistically, which is a matter of taste) for every statement in a proof to follow from previous statement, and not depend on future statements via "if and only ifs".

I will also note that the proof presented above is very elementary—it doesn't rely on any deep theorems. It is, however, somewhat tedious. More powerful theorems give quicker, more elegant proofs. For example, El31's proof, via the mean value theorem, is quite slick. However, when first learning a topic, I think that one should focus on finding an elementary proof first: find such a proof first. Then, if need be, look for something more elegant.

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The important principle here is that if $a,b$ are nonnegative real numbers, then $a<b$ iff $a^2<b^2.$ This property is used all the time. If we grant this, then all we need to show is that

$$(\sqrt{1+x})^2 < (1+x/2)^2,$$

which is the same as saying $1+x < 1+x + x^2/4.$ That is obviously true, and we're done.

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Observe that : $$ \left(\forall x>0\right),\ \sqrt{1+x}-1-\frac{x}{2}=-\frac{x^{2}}{4}\int_{0}^{1}{\frac{1-t}{\left(1+tx\right)\sqrt{1+tx}}\,\mathrm{d}t} < 0 $$

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    $\begingroup$ No offense meant, but this is perhaps one of the most obtuse correct answers I've seen on this site. $\endgroup$ – jawheele Mar 30 at 5:51

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