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I am practicing some homework and I'm stumped.

The question asks you to prove that

$x \in Z^+, y \in Z^+$

$\frac xy + \frac yx \ge 2$

So I started by proving that this is true when x and y have the same parity, but I'm not sure how to proceed when x and y have opposite partiy

This is my proof so far for opposite parity

$x,y \in Z^+ $ $|$ $x \gt 0,$ $y \gt 0$. Let x be even $(x=2a, $ $ a \in Z^+)$ and y be odd $(y=2b+1, $ $b \in Z^+)$. Then,

$\frac xy + \frac yx \ge 2$

$\frac {2a}{2b+1} + \frac {2b+1}{2a} \ge 2$

$\frac {2b^2 + 4a^2 + 4a + 1}{2b(2a+1)} \ge 2$

$4b^2 + 4a^2 +4a + 1 \ge 4b(2a+1)$

$4b^2 + 4a^2 + 4a +1 \ge 8ab + 4b$

$4b^2 - 4b + (2a + 1)^2 \ge 8ab$

$(2b-1)^2 + 1 + (2a+1)^2 \ge 8ab$

I feel like this is the not the correct way to go about proving it, but I can't think of a better way to do it. Does anyone have any suggestions? Just hints please, not a solution.

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    $\begingroup$ Hint: Since you know that x and y are positive. you can multiply the inequality by xy. $\endgroup$ – Sorfosh Oct 24 '15 at 5:19
  • $\begingroup$ @Sorfosh This helps give intuition as to why it is true, however in any correctly worded proof the desired inequality is the last line, not the first. $\endgroup$ – JMoravitz Oct 24 '15 at 5:20
  • $\begingroup$ How do you prove the above inequality is true when x and y have same parity? $\endgroup$ – Dragonemperor42 Oct 24 '15 at 5:27
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    $\begingroup$ We know that $a+\frac1a\ge2$ for any $a>0$. See here (and maybe other posts linked to that question might be of interest, too). $\endgroup$ – Martin Sleziak Oct 24 '15 at 7:37
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Hint: Consider multiplying both sides of the inequality by $xy$, which is positive. Then, we have: $$\frac xy + \frac yx \ge 2\\ x^2+y^2\ge 2xy $$ Can you go on?

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    $\begingroup$ I will point out as I did in the comments on the original post, that proving that if we assume the desired inequality is true leads to a tautology, this is not the same as proving that the desired inequality is actually true. For trivial counterexample, $5\leq 3 \Rightarrow 5\cdot 0 \leq 3\cdot 0\Rightarrow 0\leq 0$ and we know that $0\leq 0$ is true, however this doesn't imply that $5\leq 3$. A correctly worded proof will have the desired inequality at the end, not at the beginning of the proof. $\endgroup$ – JMoravitz Oct 24 '15 at 5:49
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    $\begingroup$ @JMoravitz I agree in a sense with you, but in that specific case we move through "$\iff$" and not just implications. $\endgroup$ – thanasissdr Oct 24 '15 at 5:53
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Clearly $(x-y)^2 \geq 0$

So $x^2-2xy+y^2 \geq 0 \Rightarrow x^2+y^2 \geq 2xy $

Since $x $ and $y$ are positive integers , $x \neq 0 $ and $y \neq 0$.

Thus we can divide by $xy$.

So we have $\frac{x^2+y^2}{xy} \geq \frac {2xy}{xy} \Rightarrow \frac {x}{y}+\frac{y}{x} \geq 2$.

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    $\begingroup$ Clear, concise +1 $\endgroup$ – user253055 Oct 24 '15 at 5:23
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    $\begingroup$ As the OP stated "Just hints please, not a solution.", I recommend putting at least half of this in spoiler by using >! $\endgroup$ – JMoravitz Oct 24 '15 at 5:23
  • $\begingroup$ @JMoravitz I agree on that matter, but everybody seems to be already giving full solutions anyway. $\endgroup$ – user253055 Oct 24 '15 at 5:24
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$$(x-y)^2\geq0$$

$$\begin{array}{l}x^2-2xy+y^2\geq0\\x^2+y^2\geq2xy\\\frac{x^2+y^2}{xy}\geq2\\\frac{x^2}{xy}+\frac{y^2}{xy}\geq2\\\frac{x}{y}+\frac{y}{x}\geq2\\QED\end{array}$$

This will work for $x,y\in\mathbb{R}\backslash\{0\}$

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  • $\begingroup$ As the OP stated "Just hints please, not a solution.", I recommend putting at least half of this in spoiler by using >! $\endgroup$ – JMoravitz Oct 24 '15 at 5:23
  • $\begingroup$ Hmm >! didn't work right. Let me try again. $\endgroup$ – Ian Miller Oct 24 '15 at 5:25
  • $\begingroup$ I recommend >!$\begin{array}{l} x^2-2xy+y^2\geq 0\\ ... QED\end{array}$ $\endgroup$ – JMoravitz Oct 24 '15 at 5:27
  • $\begingroup$ That didn't make the spoiler tag work properly. $\endgroup$ – Ian Miller Oct 24 '15 at 5:29
  • $\begingroup$ Thanks for the edit. Still learning a lot of SE syntax. $\endgroup$ – Ian Miller Oct 24 '15 at 5:30
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For every $a$ you have $(a-1)^2\geq 0$, so $a^2+1\geq 2a$.

If $a>0$ you have $a+\frac 1a\geq 2.$ Take $a=\frac xy$.

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If $x=y,$ then it is true.

Suppose $x>y$ then there is exists positive integer $n$ such that $x=y+n.$

This implies $\frac{x}{y}=1+\frac{n}{y}$ and $\frac{y}{x}=1-\frac{n}{x}$.
Adding these equations, we obtain \begin{align} \frac{x}{y}+ \frac{y}{x}=2+n\left(\frac{1}{y}-\frac{1}{x}\right)=2+n\frac{x-y}{xy}> 2\quad as\quad x>y. \end{align}

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Consider $f(t) = t+{1 \over t}$ on $t >0$. $f$ is convex and $f'(1) = 0$, hence $1$ is a global maximiser. Hence $f(t) \ge f(1) = 2$ for all $t > 0$.

Hence ${x \over y} + { y \over x} = f({x \over y}) \ge 2$ for all $x,y>0$.

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  • $\begingroup$ Why the downvote? It is the only downvote in this whole page. Please explain what differentiates this answer in a bad way? $\endgroup$ – copper.hat Oct 24 '15 at 16:00
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You answer is correct but you went the wrong way near the end. (Plus you have minor error in it (you swapped a and b somewhere which doesn't really change the problem).

You had: $\frac{x}{y}+\frac{y}{x}\geq2$

$\frac{2a}{2b+1}+\frac{2b+1}{2a}\geq2$

$\frac{4a^2+4b^2+4\color{red}b+1}{2\color{red}a(2\color{red}b+1)}$

$4a^2+4b^2+4b+1\geq4a(2b+1)$

$4a^2+4b^2+-8ab-4a+4b+1\geq0$

$(2a+2b-1)^2\geq0$

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Here is "another" simple method(this is essentially same as completing the square.) of proving $\frac{x}{y}+\frac{y}{x}\geq 2$ by the help of AM-GM Inequality :

Consider the set $\{\frac{x}{y},\frac{y}{x}\}.$

Applying AM-GM Inequality on these two values , we have :

$$\frac{\frac{x}{y}+\frac{y}{x}}{2} \geq \sqrt {\frac {x}{y} \times \frac{y}{x}}$$

$$\implies \frac{\frac{x}{y}+\frac{y}{x}}{2} \geq \sqrt {1}$$

$$\implies \frac{x}{y}+\frac{y}{x} \geq 1 \times 2$$

$$\implies \frac{x}{y}+\frac{y}{x} \geq 2$$

Hople this helps. :)

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  • $\begingroup$ Why a set is needed to apply AM-GM ? $\endgroup$ – A---B Feb 4 '17 at 5:11
  • $\begingroup$ Also other answers include AM-GM inequality but for a special case. $\endgroup$ – A---B Feb 4 '17 at 5:14
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    $\begingroup$ @A---B No need of any set .... It is just to "collect" all the elements on which the AM-Gm has to applied. Sometimes people get confused on which values AM-GM has been applied. To avoid that confusion, I have explicitly mentioned the elements on which AM-GM has been applied.... $\endgroup$ – Nirbhay Feb 4 '17 at 5:14
  • $\begingroup$ @A---B I have already mentioned that AM-GM for 2 values is essentially same as completing the square.... $\endgroup$ – Nirbhay Feb 4 '17 at 5:15

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