# How can I prove that $xy\leq x^2+y^2$? [closed]

How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?

• Note that you will need $(x, y) \neq (0, 0)$, to show the strict inequality. Apr 10, 2013 at 16:28
• @Maesumi In fact, you need $x\neq y$ for the strict inequality. Apr 10, 2013 at 19:09
• You can easily prove that $|xy|\leq \left(\mbox{max}\{|x| , |y| \}\right)^2$. Apr 11, 2013 at 17:50
• You also need $x$ and $y$ to be real, since it is false for $x = y = i$. Apr 15, 2013 at 14:32
• Is this a competition for the most complicated answer? 0_o What the hell is wrong with you, people? Apr 15, 2013 at 15:24

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

• This is imo one very, very nice hint. +1 Apr 10, 2013 at 16:41
• Indeed. +1. 
– Did
Apr 10, 2013 at 16:48
• Quite an elegant hint! Apr 10, 2013 at 22:28

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.

• +1 cos i like this unnecessary overkill :D I don't know if this, or the question is more elementary... Apr 10, 2013 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!
– ABC
Apr 10, 2013 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! Apr 10, 2013 at 17:06
• @exploringnet and they are. This relationship doesn't make them dependent. Apr 10, 2013 at 17:13
• @exploringnet: They're not dependent; it's not determined that $x = r/2$ until $y$ is known. Apr 11, 2013 at 7:58

Though there are quite a few proofs already given, I'd like to add a visual one. • Why so much whitespace? Apr 10, 2013 at 17:19
• @Ruslan: Sorry, I should have cropped it. I used Latex with pstricks and generated a page of output. Apr 10, 2013 at 17:54
• Thank you to whomever took the time to crop that image for me! Apr 11, 2013 at 17:13
• Nice. ${}{}{}{}$ Apr 11, 2013 at 17:50
• How do you prove $y\le x$ says that half the rectangle fits withing half the square? Jan 16, 2015 at 10:02

\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 \dfrac{x^2+y^2}{2} \\ \implies & xy\leq {x^2+y^2} \end{align} since the clearly nonnegative real $$x^2+y^2$$ clearly satisfies $$\dfrac{x^2+y^2}{2} \leq x^2+y^2$$.

• I've taken the liberty to replace the slightly confusing update by a clear argument. Dec 5, 2019 at 20:38

Another interesting answers - maybe the most interesting:

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

• just the sum of two squares --simple and elegant .. Apr 11, 2013 at 5:35
• Very nice !+1 for you. Apr 11, 2013 at 17:51

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!

• 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. Apr 11, 2013 at 7:42

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

• You'd want a $\leq$ sign instead. Apr 10, 2013 at 16:27
• @CalvinLin yes I do :D thanks Apr 10, 2013 at 16:32 $x^2+y^2=AC^2$

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

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

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

$$(x^2 - xy + y^2)(x + y) = x^3 + y^3.$$ $x^3 + y^3$ and $x + y$ have the same sign: $x \leq -y$ iff $x^3 \leq -y^3$. Therefore, $x^2 - xy + y^2 \geq 0$.

About the special case $x = -y$: it’s not really that special, but for a technical reason that goes beyond precalculus. Since non-trivial Zariski open sets are dense in the Euclidean topology, and polynomial maps are continuous between Euclidean topologies, all inequalities of the form $f(x_1, \ldots, x_n) \geq 0$ (where $f$ is a polynomial) that hold on a Zariski open set must hold everywhere, because $[0, +\infty)$ is closed and its preimage must thus contain the closure of said Zariski open set, i.e. the whole space. Therefore, such apparent special cases can be safely ignored as long as they’re Zariski closed and the inequality is not strict.

• Technically, this needs the $x=-y$ case to be treated separately. May 9, 2013 at 3:35
• @ronno Not really, because $X := \mathbb{R}^2 \setminus \{x = -y\}$ is dense in $\mathbb{R}^2$, and $[0, +\infty)$ is closed in $\mathbb{R}$, therefore its pullback along $(x, y) \mapsto x^2 - xy + y^2$ must be $\operatorname{cl}X = \mathbb{R}^2$. Of course it is too technical for a precalculus-level question. Aug 14, 2014 at 20:18

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 $1$.

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

Just look at this picture below: • 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 $x, y \geq 0$.) Apr 11, 2013 at 8:02
• You can assume that $x \leq y$ WOLOG. If $x,y \geq 0$, then treat $a$ as $x$ and $b$ as $y$. Jun 28, 2020 at 22:11

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

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 $|x|\ge |y|$

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

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

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

• Thanks for the down-vote. I could find a better way to prove it. May 10, 2013 at 14:18
• 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, 2013 at 15:57

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...$$x^2 + y^2 \ge xy$$

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.

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.

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 $x=0$ . $\hspace {33mm} \blacksquare$

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

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

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

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

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

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

$x^2 + y^2 > xy$

• If $xy < 0$ then $2xy < xy$ Apr 11, 2013 at 13:49

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.

Say $y\ne 0$ then divide both sides $xy\le x^2+y^2$ by $y^2$; let $z=\frac{x}{y}$; then we need $z\le 1+z^2$ or equivalently $z^2-z+1\ge 0$; completing the square gives $(z-\frac{1}{2})^2+\frac{3}{4}\ge 0$ which is clear.

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 $y>=x$.