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If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?

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13 Answers 13

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Arithmetic Mean $\ge$ Geometric Mean

$\frac{x+y}{2} \ge \sqrt{xy} $

$ \implies x + y \ge 2 $

$ \implies $ Minimum Value = $2$

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  • $\begingroup$ Very nice. Should get +5 or so, can only give +1. $\endgroup$
    – vonbrand
    Mar 1, 2013 at 14:43
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    $\begingroup$ You should add trivial fact that 2 is reachable. x^2 + 1 >= 0, but 0 isn't minimum value $\endgroup$
    – RiaD
    Mar 1, 2013 at 14:50
  • $\begingroup$ @vonbrand: We can help with that. $\endgroup$
    – Stu
    Mar 1, 2013 at 19:42
  • $\begingroup$ Hey my TA showed us this in class before! $\endgroup$ Feb 6, 2014 at 8:00
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    $\begingroup$ @Derek朕會功夫 Nice ! Will definitely help you to solve many inequalities in future ! $\endgroup$
    – lsp
    Feb 6, 2014 at 8:32
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$x+y=x+y-2\sqrt{xy}+2\sqrt{xy}=(\sqrt{x}-\sqrt{y})^2+2\ge2$

The value will be attained when $(\sqrt{x}-\sqrt{y})^2=0\Rightarrow \sqrt{x}=\sqrt{y}\Rightarrow x=y=1$

Please note that this is the basis for the inequality of A.M.$\ge$ G.M.

If $x,y\ge 0$ then we know that $(\sqrt{x}-\sqrt{y})^2\ge0$

$\Rightarrow x+y-2\sqrt{xy}\ge 0$

$\Rightarrow x+y\ge 2\sqrt{xy}$

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With $x\cdot y=1$ you know that $x=\frac{1}{y}$ so $$x+y=x+\frac{1}{x}$$ Wlog you can say now $x\geq 1 $, try to find the minimum now.

Another way is for x>0 we know that $$\frac{(x-1)^2 }{x}\geq 0$$ on the other hand $$\frac{(x-1)^2}{x} = \frac{x^2}{x} - \frac{2x}{x} +\frac{1}{x} =x-2 +\frac{1}{x} \geq 0$$ This is equivalent to $$x+\frac{1}{x}\geq 2$$ and as $1+1=2$ we know that 2 is the minimum.

I guess I get some downvotes for the following, but maybe it gives you some hope. If you don't like adding a smart zero (which is a bit unintuitive), you can still use advanced calculus methods. You try to minimize the function $f:\mathbb{R}^2\rightarrow \mathbb{R}; \ (x,y)\mapsto x+y$ with the side condition $x\cdot y = 1$.

Now you use a lagrangian multiplier and minimizing the function $$g(x,y)=x+y+\lambda (xy-1)$$ Now you take partial derivatives, which must be zero for a minimum \begin{align*} \frac{\partial g}{\partial x} &= 1+ \lambda y\\ \frac{\partial g}{\partial y} &= 1+ \lambda x\\ \frac{\partial g}{\partial \lambda} &= xy-1 \end{align*} Now you only need to solve the system \begin{align*} 0&=1+\lambda y\\ 0&= 1+\lambda x \\ 0&=xy - 1 \end{align*}

As Winston mentioned you could just take the derivative of $\displaystyle x+\frac{1}{x}$ which is $$1-\frac{1}{x^2}$$ So you need to solve the equation $$1-\frac{1}{x^2}=0 \iff 1=\frac{1}{x^2} \iff x^2=1$$

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    $\begingroup$ Nice (+1), but that is cheating for the tag... $\endgroup$
    – vonbrand
    Mar 1, 2013 at 14:42
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    $\begingroup$ Why would you go for advanced calc methods rather then taking the derivative of x + 1/x? $\endgroup$ Mar 1, 2013 at 15:27
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Note that $y=1/x$ and $$\left(x+\frac{1}{x}\right)^2=\left( x-\frac{1}{x}\right)^2+4\cdot x\cdot \frac{1}{x}\geq4$$ Hence $x+y\geq 2$. Also for $x=y=1$, you have $x+y=2$, hence $2$ indeed the minimum value.

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The above answer does it totally.

However you can visualize this problem graphically. Imagine the three axes $x,y,z$ and the first quadrant part of the curve $xy=1$ drawn on the $xy$ plane. The function $f(x,y)=x+y$ is plotted on the $z$-axis. The graph of this function looks like a plane which is slanted 45% to the $xy$-plane and cuts it on the line $135^\circ$ to the $x$-axis. The minimum of this function with the constraint occurs where the curve $xy=1$ comes closest to the origin. You can see this by imagining the $z=x+y$ plane being parallely translated up and down.

Then it is clear that since a line of slope $135^\circ$ is tangent to the curve $xy=1$, this point of incidence will give you the minimum of $f$.

This point of incidence is $x=1,y=1$. and hence the minimum is $2$. And this picture also makes it clear that the minimum is attained.

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  • $\begingroup$ +1, good visual answer. (BTW note that "the above answer" will vary over time as answers are added and voted on. So you may want to say whose answer you're referring to.) $\endgroup$
    – LarsH
    Mar 1, 2013 at 20:02
  • $\begingroup$ oops I hadn't noticed it. I am using this floating answer forum for the first time. I was referring to Abhra's answer at that time .. now some more nice answers has emerged. Thanks for your appreciation though. $\endgroup$
    – magguu
    Mar 1, 2013 at 20:55
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$$x+y=\frac{x^2+1}{x}=\frac{x^2-2x+1}{x}+2 \geq0+ 2$$

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This question has had a lot of answers, but none use what I would consider to be the easiest proof, so here's a general way to solve the problem:

$xy = 1$ can be re-written as $y = {1\over x}$, and putting that back into $x+y$ we get:

$$x+y=x+{1\over x}=x+x^{-1}$$

which differentiates to:

$${d\over dx} (x+x^{-1})=1-x^{-2}$$

Now at a turning point this will be 0. That's at $x=\pm 1$. So the two turning points are at $x=-1,y=-1$ and $x=1,y=1$.

You already restricted the solution to positive numbers (them being real makes no difference), but for completeness we still need to look at the edge cases of $x=\inf$ and $x=0$ (which both produce infinite results).

Instead of taking edge cases, you could also show that this is a minimum (rather than maximum or point of inflection) by taking the second differential and checking the sign: $${d^2\over dx^2} (x+x^{-1})=2x^{-3}$$

At $x=1$, this is $2$. Since this is greater than 0, we know it is a minimum. Note that in this case it is the only turning point within the range, so must be a global minimum, but in general you would have to check every turning point.

So your minimum is $2$, at $x=1, y=1$.

Some references if you're new to derivatives:

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  • $\begingroup$ I noticed that @Dominic Michaelis mentions this method briefly in his comment, but he doesn't prove that it is a global minimum in this way. $\endgroup$
    – Dave
    Mar 1, 2013 at 23:36
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    $\begingroup$ The question is tagged algebra-precalculus, so I'd assume a solution without derivatives is called for. $\endgroup$ Mar 1, 2013 at 23:37
  • $\begingroup$ Ah, good point, I missed that (although calculus is certainly the easiest way to do this, and this is very simple-level calculus) $\endgroup$
    – Dave
    Mar 2, 2013 at 0:20
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Here is another approach. We have $ x, y > 0 $ and

$$ xy=1 \implies y=\frac{1}{x}. $$

Substituting $ y=\frac{1}{x} $ in the equation $f(x,y)=x+y$ yields

$$ F(x)= x+\frac{1}{x}.$$

Now, we have a function in one variable which can be minimized using techniques from univariate calculus as

$$ F(x)= x+\frac{1}{x} \implies F'(x)=1-\frac{1}{x^2}=0\implies x = 1>0. $$

$$ \implies F''(1)=2>0 $$

which means $x=1$ is a minima for the function $F(x)$ and the minima is $F(1)=2.$

So the minimum of the function $f(x,y)$ attains at the point $ (x,y)=(1,1) $.

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    $\begingroup$ Note that the question is tagged algebra-precalculus. $\endgroup$ Mar 1, 2013 at 23:38
  • $\begingroup$ @GerryMyerson: I asked a question to the OP and I deleted asking where this problem came from before I posted my answer, but no answer. I think they study applications of derivative in precalculus course. $\endgroup$ Mar 1, 2013 at 23:51
  • $\begingroup$ @GerryMyerson: Did you check Dominic's answer? $\endgroup$ Mar 1, 2013 at 23:54
  • $\begingroup$ I saw Dominic's answer, and I saw that others had already commented on its level. What's your point? Anyway, I take "precalculus" to mean "no calculus allowed, in particular, no use of derivatives." I could be wrong, but that's my interpretation. $\endgroup$ Mar 2, 2013 at 5:19
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I agree with magguu about visualizing the problem graphically, but I'd do the visualization in two dimensions rather than three. In the $x$-$y$ plane, imagine the graph of $xy=1$ (a hyperbola) and imagine the level lines of $x+y$ (lines sloping downward, making $45^\circ$ angles with the axes). In more detail, start with the level line through the origin ($xy=0$), which misses the hyperbola entirely, and imagine gradually moving this line to larger values of $x+y$, until it first touches the hyperbola. By symmetry between $x$ and $y$ (and convexity of the hyperbola), this happens at the point on the hyperbola where $x=y$, i.e., at the point $(1,1)$.

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$$(\sqrt x-\sqrt y)^2\geq 0$$ $$x-2\sqrt{xy}+y\geq 0$$ $$x+y\geq 2\sqrt{xy}$$ $$x+y\geq 2\cdot 1$$ $$x+y\geq 2$$

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If for $c>0, xy=c^2\implies y=\frac{c^2}x\implies x+y=x+\frac{c^2}x=z$(say)

So, $x^2-zx+c^2=0$

As this is a quadratic equation in $x$ and $x$ is real,

the discriminant $(-z)^2-4\cdot1\cdot c^2\ge 0\implies z^2\ge4c^2\implies $ either $z\ge2c$ or $z\le -2c$

As $x,y>0,x+y>0$

So, $x+y=x+\frac1x=z\ge2c$

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Consider a rectangle with dimension (x, y) of unit area, the half perimeter is defined as $a+b$

a square is a one and only one special form of rectangle with equal dimension should either be a maxima or minima

Now lets calculate the half perimeter of a square

$x.y = 1$

as $x = y$ we have $x.x = 1$

so $x = 1$ which gives us $x + y = 2$

Consider any one rectangular figure

$x = 2$ and $y = \frac{1}{2}$ this gives the perimeter as

$x + y = 2 + 1/2 = 2.5$

$2.5(rectangle) > 2 (square)$

which makes us to believe

a square will have the minimum perimeter

thus we can conclude

the minimum value of $x + y$ is $2$

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as $xy=1$;

so, $y=1/x$;

as we know that $(x+y)^2 = x^2+y^2+2xy$;

$(x+y)^2 = x^2 = 1/x^2 +2$;

so, $x+y = \sqrt{x^2 + \frac{1}{x^2} + 2}$ to be minimum, compare RHS by 0; you will get $x=-1$; but as $x$ is a +ve real number, so, $x$ is minimum at 1; so $x+y =2$;

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