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How can I prove that $xy\leq x^2+y^2$?

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you can even prove $|2xy|<x^2+y^2$ by using some basic identities. –  Maesumi Apr 10 '13 at 16:27
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Note that you will need $(x, y) \neq (0, 0)$, to show the strict inequality. –  Calvin Lin Apr 10 '13 at 16:28
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You can easily prove that $|xy|\leq \left(\mbox{max}\{|x| , |y| \}\right)^2$. –  N. S. Apr 11 '13 at 17:50
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You also need $x$ and $y$ to be real, since it is false for $x = y = i$. –  marty cohen Apr 15 '13 at 14:32
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Is this a competition for the most complicated answer? 0_o What the hell is wrong with you, people? –  Alexei Averchenko Apr 15 '13 at 15:24
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26 Answers

up vote 18 down vote accepted

\begin{align} 0\leq (x-y)^2 \implies & 0\leq x^2-2xy +y^2 \\ \implies & 2xy\leq x^2+y^2 \\ \implies & xy\leq {x^2+y^2} \end{align}

Update. This proves I'm not assuming that $xy \leq 2xy$! Clearly this is only valid for non-negative $xy$. Note that if $xy\leq 0 $ if, only if, $2xy\leq 0$.

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If you remove the divisions by $2$ and write $xy\leq 2xy$ instead, I will happily upvote. The only price to pay is to assume $xy\geq 0$. But if $xy<0$, there is nothing to prove. But don't feel any pressure! –  1015 Apr 10 '13 at 17:04
    
@julien Nice sugestion. –  Elias Apr 10 '13 at 23:32
    
I kept my word, +1. –  1015 Apr 10 '13 at 23:49
    
@Calvin Lin , because you gave me a downvote? –  Elias Apr 11 '13 at 1:04
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Maybe that's because you didn't say at the beginning that you assumed $xy\geq 0$. For otherwise $xy\leq 2xy$ is no true. Also, $xy\leq 2xy$ does not explicitly appear. Sorry if you got one downvote because of me... I still think your answer is the best here. –  1015 Apr 11 '13 at 1:08
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$$x^2+y^2-xy=\frac{x^2}{2}+\frac{y^2}{2}+\frac{(x-y)^2}{2}$$

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This is imo one very, very nice hint. +1 –  DonAntonio Apr 10 '13 at 16:41
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Indeed. +1. $ $ –  Did Apr 10 '13 at 16:48
    
Quite an elegant hint! –  Barranka Apr 10 '13 at 22:28
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Use polar coordinates: $$x=r \cos \theta,y=r \sin \theta $$

Your inequality becomes $$r^2 \cos \theta \sin \theta \leq r^2 $$

which is pretty much trivial.

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+1 cos i like this unnecessary overkill :D I don't know if this, or the question is more elementary... –  Lost1 Apr 10 '13 at 16:47
    
But you make both $x,y$ bound . ie. less than $r$ and even related to each other . $x^2+y^2=r^2$ .They become the co-ordinates of a circle! –  Mr.ØØ7 Apr 10 '13 at 16:49
    
@exploringnet I don't see a problem. Any point in the plane has its polar coordinates. The radius $r$ is definitely not constant! –  user1337 Apr 10 '13 at 17:06
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@exploringnet and they are. This relationship doesn't make them dependent. –  user1337 Apr 10 '13 at 17:13
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@exploringnet: They're not dependent; it's not determined that $x = r/2$ until $y$ is known. –  LarsH Apr 11 '13 at 7:58
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Though there are quite a few proofs already given, I'd like to add a visual one.

picture

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Why so much whitespace? –  Ruslan Apr 10 '13 at 17:19
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@Ruslan: Sorry, I should have cropped it. I used Latex with pstricks and generated a page of output. –  Shaun Ault Apr 10 '13 at 17:54
    
Thank you to whomever took the time to crop that image for me! –  Shaun Ault Apr 11 '13 at 17:13
    
Nice. ${}{}{}{}$ –  The Chaz 2.0 Apr 11 '13 at 17:50
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another interesting answers - maybe the most interesting:

$$x^2+y^2-xy=\left(x-\frac{y}{2}\right)^2+\frac{3y^2}{4} \geq 0.$$

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just the sum of two squares --simple and elegant .. –  Halil Duru Apr 11 '13 at 5:35
    
Very nice !+1 for you. –  Elias Apr 11 '13 at 17:51
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For $xy<0$, this is trivial

You can do better by proving $2xy \leq x^2 + y^2$. Move $2xy$ to the right handside and factorise

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You'd want a $\leq$ sign instead. –  Calvin Lin Apr 10 '13 at 16:27
    
@CalvinLin yes I do :D thanks –  Lost1 Apr 10 '13 at 16:32
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enter image description here

$x^2+y^2=AC^2$

$\sin \angle BAC= \dfrac{BC}{AC}$

$\sin \angle BCA= \dfrac{AB}{AC}$

$\sin \angle BCA \cdot \sin \angle BAC <1 \implies \dfrac{xy}{x^2+y^2} <1$.

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A proof with only words: assume $x$ and $y$ are positive, if one is negative this is trivial. Then one of $x$ or $y$ is smallest, the other largest. Assume $y$ is the largest (else switch). Then $xy$ is clearly less than $y^2$, and adding $x^2$ doesn't change that!

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I like proofs with only words. :-) This answer has been the most accessible one for me. You might want to mention the case where $x$ and $y$ are both negative, though. –  LarsH Apr 11 '13 at 7:42
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$$(x^2 - xy + y^2)(x + y) = x^3 + y^3.$$ $x^3 + y^3$ and $x + y$ have the same sign: $x \leq -y$ iff $x^3 \leq -y^3$. Therefore, $x^2 - xy + y^2 \geq 0$.

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Technically, this needs the $x=-y$ case to be treated separately. –  ronno May 9 '13 at 3:35
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If $xy$ is negative, then the statement is obvious since $xy < 0 \leq x^2 + y^2$.

Otherwise, $xy$ is non-negative, and we can show that $xy \leq 2xy \leq x^2+y^2$, where the latter follows from the trivial inequality $(x-y)^2 \geq 0 $.

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Just look at this picture below:

Geometry proof

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Trying to make sure I understand how $a, b$ relate to $x, y$. Are we to assume that $b = max(x, y)$ and $a = min(x, y)$? (And that $x, y \geq 0$.) –  LarsH Apr 11 '13 at 8:02
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Another one - if $xy \le 0$ it is trivial, so let $xy > 0$. Then we have $$ 1 \le \dfrac{x}{y}+\dfrac{y}{x}$$

Now for any positive number, either it or its reciprocal must exceed $1$, unless both are $1$.

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A different approach, $$A.M[x_i^n]\ge(A.M[x_i])^n\ge GM[x_1]^n$$

ie. AM of terms with $n$th power is greater than $n$th power of AM ie. greater than $n$th power of GM . So,$$\dfrac{x^2+y^2}2\ge(\dfrac{x+y}2)^2\ge(xy)$$ and so,$$(x^2+y^2)\ge xy$$

And for $xy<0$ the inequality is obvious.

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$$x^2+y^2-xy=\frac{(2x-y)^2+3y^2}4=\frac{(2x-y)^2+(\sqrt3y)^2}4$$

Now, the square of any real numbers is $\ge0$

So, $(2x-y)^2+(\sqrt3y)^2\ge0,$ the equality occurs if each $=0$

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Thanks for the down-vote. I could find a better way to prove it. –  lab bhattacharjee May 10 '13 at 14:18
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Dear lab bhattacharjee, to be honest, I should admit that it is me who downvoted your answer earlier. I did the downvote not because your answer is not good, but personally I don't encourage users with high reputation like you to pay much attention to trivial questions like this one. In my opinion, supporting such questions will diminish the quality of this website. By the way, I also downvoted lots of other answers(to this/other post) provided by users with high reputation for the same reason. In view of your serious attitude to downvotes, I have to cancel my downvote. Sorry for bothering. –  23rd May 10 '13 at 15:57
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Assume $x$ and $y$ are positive. Draw the right triangle with vertices $(0,0), (x,0), (x,y)$. Draw the circle around $(0,0)$ that circumscribes the triangle. Reflect the triangle in the $x$-axis. Reflect the two triangles in the $y$-axis too, for a better result.

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Intuitive Approach

if $|x|\le |y|$ then we have $$x\cdot y \le y\cdot y \le y.y + x\cdot x$$ $$\Rightarrow x\cdot y \le x^2+y^2\tag1$$ Using Symmetry $(1)$ holds if $|x|\ge |y|$

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This is clearly a consequence of the concavity of the logarithm and the monotonicity of the exponential.

Since $xy\leq |x||y|$, we can assume that $x >0$ and $y > 0$. Then we have $$ xy = \exp(\ln(xy)) = \exp(\ln x + \ln y) = \exp(\frac12\ln x^2 + \frac12 \ln y^2) \leq \exp(\ln(\frac{x^2}2 + \frac{y^2}2)) =\frac{x^2}2 + \frac{y^2}2 \leq x^2 + y^2. $$ QED

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Technique 1:$$\sqrt{x^2y^2} = xy \le \dfrac{x^2 + y^2}{2}\le x^2 + y^2$$ Technique 2:$$(x - y)^2 \ge 0 \implies x^2 +y^2 - 2xy \ge 0 \implies x^2 + y^2 \ge 2xy\ge xy$$ Technique 3 (my favorite): There's a statement $\dfrac{y}{x} + \dfrac{x}{y} \ge 2$ with many classical proofs (which I'd not state here). We can write the inequality as follows:$$\dfrac{y}{x}+\dfrac{x}{y} \ge 1$$Divide both sides by $xy$.$$\dfrac{1}{x^2} + \dfrac{1}{y^2} \ge \dfrac{1}{xy}$$Rewrite.$$\dfrac{x^2 + y^2}{x^2y^2} \ge \dfrac{xy}{x^2y^2}$$And finally...$$x^2 + y^2 \ge xy$$

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Let $y=kx$ then $ x xk \leq x^2 + x^2k^2$ since $k\leq k^2+1$ [in view of $0\leq (k-1/2)^2+3/4] $

And the ineq. clearly holds when $x=0$ . $\hspace {33mm} \blacksquare$

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$$(x-y)^2=x^2+y^2-2xy\\ (x-y)^2+2xy=x^2+y^2\\ 2xy\leq x^2+y^2$$ Therefore $xy\leq x^2+y^2$. Hence the proof.

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First, $$ a^2+3b^2\geq 0 $$ Now, $$ a^2+2ab+b^2+a^2-2ab+b^2\geq a^2-b^2 $$ Thus, $$ (a+b)^2+(a-b)^2\geq (a+b)(a-b) $$ Let $a+b=x, a-b=y$, i.e., $a=\frac{x+y}{2}, b=\frac{x-y}{2}$ which is one-to-one correspondence between $(x,y)$ and $(a,b)$, then, $$ x^2+y^2\geq xy $$ Q.E.D

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$(x - y)^2 \geq 0$

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

$x^2 + y^2 \geq 2xy > xy$

upate Happy Green Kid is right below, the case for xy < 0 needs to be considered:

so, continuing from:

$x^2 + y^2 \geq 2xy$

if $xy > 0:$

$x^2 + y^2 \geq 2xy > xy$

else

$x^2 + y^2 > 0 > xy$

in both cases

$x^2 + y^2 > xy$

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If $xy < 0$ then $2xy < xy$ –  Happy Green Kid Naps Apr 11 '13 at 13:49
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If xy is negative, then it is trivial. If both x and y are negative, then let x := -x and y := -y. This means that we can safely assume that both x and y are positive.

Square each sides yields

$$x^2y^2 \le x^4 + 2x^2y^2 + y^4$$

, which trivially holds.

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Let $y>=x. xy<=x^2+y^2$. Divide both sides by $y$, and you get $x<=(x^2/y)+y$. If $x$ and $y$ are both positive numbers, this is clearly true because we said from the beginning that $y>=x$.

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My viral 5 cents:

\begin{equation} xy\le x^2+y^2 \end{equation} rearrange it \begin{equation} 0\le x^2+y^2 - xy \implies \end{equation} \begin{equation} 0\le \frac{x^2 + y^2 + (x-y)^2}{2} \implies \end{equation} \begin{equation} 0\le x^2 + y^2 + (x-y)^2, \forall x,y \in \mathbb{R} \end{equation} finally, let's prove $$ a^2 \ge 0, \forall a \in \mathbb{R} $$ $\mathbb{R} $ is unitary ring $ \implies $ has an inverse element: $$ a + (-a) = 0 \implies a = -(-a) $$ $$ 0 = 0 \cdot b = (a + (-a)) \cdot b = a \cdot b + (-a) \cdot b = 0\implies $$ $a \cdot b$ and $ (-a) \cdot b $ are inverse to each other, but in turn, inverse element for $ a \cdot b $ is $ -(a \cdot b) \implies $ $$ (-a) \cdot b = -(a \cdot b) $$ or $$ (-a) \cdot (-a) = -(a \cdot (-a)) = -(-a) \cdot a = a \cdot a $$ thus $$ x \cdot x + y \cdot y + (x - y) \cdot (x - y) $$ let $z = x -y$ and we will have $$ x \cdot x + y \cdot y + z \cdot z = x^2 + y^2 + z^2 \ge 0, \forall x,y,z \in \mathbb{R} \implies \\ \\x^2+y^2 + (x-y)^2 \ge 0, \forall x,y \in \mathbb{R} \implies $$ \begin{equation} xy\le x^2+y^2, \forall x,y \in \mathbb{R} \end{equation} $$ Q.E.D. $$

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for downvote: if it isn't a plain racism :) please show me my faults to learn –  rook Apr 12 '13 at 10:59
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(-1) since this is just wrong: $a^2 \geq b^2$ does not imply $a \geq b$. –  TMM Apr 15 '13 at 13:50
    
@TMM, you're right, possibly there is another kind of problem - is to prove that the squaring in certain circumstances didn't broke an inequality sign –  rook Apr 15 '13 at 14:48
    
This is really not different (now) than the earlier highly upvoted answers. –  mixedmath Apr 16 '13 at 17:35
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\begin{equation} xy\le x^2+y^2 \end{equation} Move xy to the right \begin{equation} \implies 0 \le x^2 - xy + y^2 \end{equation} This becomes the prove that \begin{equation} x^2 - xy + y^2 \end{equation} is always positive AKA greater or equal to zero we can rewrite it as \begin{equation} 0 \le x^2 - 2(\dfrac{1}{2} xy) + y^2 \end{equation} Which simly means that we can write it as: \begin{equation} 0 \le x^2 - 2(\dfrac{1}{2} xy) + y^2 \end{equation} Our equation has the form: \begin{equation} a^2 - 2(ab) + b^2 = (a - b)^2 \end{equation} By factorization we get: \begin{equation} 0 \le 2\left(\sqrt{\dfrac{1}{2}}(x) - \sqrt{\dfrac{1}{2}}(y)\right)^2 \end{equation}

Since \begin{equation} 2 \end{equation} is positive \begin{equation} 2\left(\sqrt{\dfrac{1}{2}}(x) - \sqrt{\dfrac{1}{2}}(y)\right)^2 \end{equation} is positive as the square of a number so: \begin{equation} 2\left(\sqrt{\dfrac{1}{2}}(x) - \sqrt{\dfrac{1}{2}}(y)\right)^2 \end{equation} Can only be positive meaning: \begin{equation} 0 \le 2\left(\sqrt{\dfrac{1}{2}}(x) - \sqrt{\dfrac{1}{2}}(y)\right)^2 \end{equation} Is always true.

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