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Which real numbers satisfy $\frac{x^2-2x+3}{\sqrt{x^2-2x+2}}\geq 2 $ ?

A. The set of all real numbers

B. The set of all integers

C. The set of all positive real numbers

D. The set of all rational numbers

My solution is:

We consider, $x^2-2x+2=u$ then, $\frac{x^2-2x+3}{\sqrt{x^2-2x+2}}=\frac{u+1}{\sqrt{u}}\geq 2\implies (u+1)^2\geq 4u\implies u^2+1-2u\geq 0\implies (u-1)^2\geq 0,$ which is true. Hence, any $x\in\Bbb R$ satisfies $\frac{x^2-2x+3}{\sqrt{x^2-2x+2}}\geq 2.$ Also, as $u$ is always greater than $1$, so we can say option $A$ as our answer.

Is this solution correct?

Arthur
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    "it yields a huge expression" Try to write it as $\frac{u^2+1}{u} \ge 2$ instead. – dxiv Apr 22 '23 at 05:15

3 Answers3

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First of all, we should verify that the denominator is well-defined. The quadratic $x^2-2x+2$ has its vertex at $x = -(-2)/(2) = 1$, and $1^2 - 2(1) = 2 = 1$, so we see $x^2 - 2x + 2 \geq 1$ for all $x \in \mathbb{R}$.

Now notice that we have $$ \frac{x^2-2x+3}{\sqrt{x^2-2x+2}} = \frac{x^2-2x+2}{\sqrt{x^2-2x+2}} + \frac{1}{\sqrt{x^2-2x+2}} = \sqrt{x^2-2x+2} +\frac{1}{\sqrt{x^2-2x+2}}. $$ But since we know $\sqrt{x^2-2x+2} \geq 1$, we can look at the equivalent problem of showing $u + 1/u \geq 2$ for any $u \geq 1$.

We have the following: $$ (u-1)^2 \geq 0 \implies u^2-2u+1 \geq 0 \implies u^2+1 \geq 2u \implies u + \frac1u \geq 2 $$ since $u > 0$. Therefore, we conclude $$ \frac{x^2-2x+3}{\sqrt{x^2-2x+2}} \geq 2 $$ for all real numbers $x$.

Nick F
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  • I 'hv edited OP mind taking a look at it? – Arthur Apr 22 '23 at 05:27
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    @Arthur Your edit looks mostly good to me. You should still verify at least that your $u$ is indeed positive, so that you can take the square roots as indicated. That's why I showed that $x^2 - 2x + 2 \geq 1$. – Nick F Apr 22 '23 at 05:36
  • @Nick F Thanks a lot! I do get it now... – Arthur Apr 22 '23 at 05:39
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Alternatively, since this is a multiple choice test you can simply plug in $-\pi$ (with the knowledge that $3<\pi<4$) and see that

$$\frac{(-\pi)^2+2\pi+3}{\sqrt{(-\pi)^2+2\pi+2}}=\frac{\pi^2+2\pi+3}{\sqrt{\pi^2+2\pi+2}}$$

$$>\frac{3^2+2\cdot 3+3}{\sqrt{4^2+2\cdot 4+2}}=\frac{18}{\sqrt{26}}>\frac{18}{\sqrt{36}}=3>2$$

Then the only possible answer that works is $A$.

QC_QAOA
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Completing the square, we see that both the numerator and the radical within the denominator are always positive, so we'll be able to square the expression when convenient without introducing extraneous roots.

\begin{align} \frac{(x-1)^2+2}{\sqrt{(x-1)^2+1}} \geq 2 &\iff [(x-1)^2+2]^2 \geq 2[(x-1)^2+1] \\ &\iff (x-1)^4 +4(x-1)^2 + 4 \geq 2(x-1)^2 + 2 \\ &\iff (x-1)^4+2(x-1)^2 + 2 \geq 0 \\ &\iff [(x-1)^2 + 1]^2 +1 \geq 0. \end{align}

The last line is obtained by again completing the square. This final inequality obviously holds for all $x \in \Bbb R$.

As noted above, since this is a multiple choice question and "none of the above" isn't an option, you know from the outset that the answer must be an interval so you only need to test $0$ to determine whether that interval is larger than $\Bbb R^+$.

Robert Shore
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