# How to prove $\frac xy + \frac yx \ge 2$

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.

• Hint: Since you know that x and y are positive. you can multiply the inequality by xy. Commented Oct 24, 2015 at 5:19
• @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. Commented Oct 24, 2015 at 5:20
• How do you prove the above inequality is true when x and y have same parity? Commented Oct 24, 2015 at 5:27
• 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). Commented Oct 24, 2015 at 7:37

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

• Clear, concise +1
– user253055
Commented Oct 24, 2015 at 5:23
• As the OP stated "Just hints please, not a solution.", I recommend putting at least half of this in spoiler by using >! Commented Oct 24, 2015 at 5:23
• @JMoravitz I agree on that matter, but everybody seems to be already giving full solutions anyway.
– user253055
Commented Oct 24, 2015 at 5:24

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?

• 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. Commented Oct 24, 2015 at 5:49
• @JMoravitz I agree in a sense with you, but in that specific case we move through "$\iff$" and not just implications. Commented Oct 24, 2015 at 5:53

$$(x-y)^2\geq0$$

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

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

• As the OP stated "Just hints please, not a solution.", I recommend putting at least half of this in spoiler by using >! Commented Oct 24, 2015 at 5:23
• Hmm >! didn't work right. Let me try again. Commented Oct 24, 2015 at 5:25
• I recommend >!$\begin{array}{l} x^2-2xy+y^2\geq 0\\ ... QED\end{array}$ Commented Oct 24, 2015 at 5:27
• That didn't make the spoiler tag work properly. Commented Oct 24, 2015 at 5:29
• Thanks for the edit. Still learning a lot of SE syntax. Commented Oct 24, 2015 at 5:30

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 $a=\frac xy$.

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 $\frac{y}{x}=1-\frac{n}{x}$.

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 $x,y>0$.

• Why the downvote? It is the only downvote in this whole page. Please explain what differentiates this answer in a bad way? Commented Oct 24, 2015 at 16:00

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

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

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

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

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

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

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. :)

• Why a set is needed to apply AM-GM ?
– user312097
Commented Feb 4, 2017 at 5:11
• Also other answers include AM-GM inequality but for a special case.
– user312097
Commented Feb 4, 2017 at 5:14
• @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....
– user399078
Commented Feb 4, 2017 at 5:14
• @A---B I have already mentioned that AM-GM for 2 values is essentially same as completing the square....
– user399078
Commented Feb 4, 2017 at 5:15

Without loss of generality, we assume $$t:=\frac{x}{y}\geq 1,$$ we see that the $$t+\frac{1}{t}$$ has its derivative non-negative, so $$t+\frac{1}{t}\geq 2.$$

$$\frac xy + \frac yx = x\cdot \frac 1 y + y \cdot \frac 1x\ge x\cdot \frac 1 x + y \cdot \frac 1y=1+1=2$$