Find all real numbers $x,y,z\geq 1$ satisfying \[\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]
Problem
Source: 2013 USAJMO #6/USAMO #4
Tags: AMC, USA(J)MO, USAJMO, parameterization, inequalities
02.05.2013 01:03
02.05.2013 01:05
DARN FORGOT TO CONSIDER EQUALITY CASE
02.05.2013 01:06
Without loss of generality, let $\min(x, y, z) = z$. Then we have to find all solutions $(x, y, z)$, where $x, y, z \ge 1$ such that \[\sqrt{z+xyz} = \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.\]Let $x-1=r^2, y-1 = s^2, z-1=t^2$ for $r, s, t \ge 0$ and $r, s, t \in \mathbb{R}$. Then\[\sqrt{(t^2+1)(1+(r^2+1)(s^2+1))} = r+s+t\]or $(rst)^2+(rs)^2+(st)^2+(tr)^2+t^2+2-2(rs+st+tr) = 0$. By AM-GM, \begin{align*}0 &= (rst)^2+(rs)^2+(st)^2+(tr)^2+t^2+2-2(rs+st+tr) \\&= rst^2\left(rs+\frac{1}{rs}\right) +(rs)^2+(st)^2+(tr)^2+2-2(rs+st+tr) \\& \ge (rs)^2+(st)^2+(tr)^2+2rst^2 + 2 - 2(rs+st+tr) \\&= (rs-1)^2+(st+tr)^2+1-2(st+tr) \\&= (rs-1)^2+(st+tr-1)^2 \ge 0\end{align*} so equality is reached when $rs = 1$ and $st+tr = 1$, or $\sqrt{x-1}\sqrt{y-1} = 1$ and $(\sqrt{x-1}+\sqrt{y-1})\sqrt{z-1} = 1$. Solving for $y$ in the former yields $y = \frac{x}{x-1}$, and squaring the latter yields \begin{align*} ((x-1)+(y-1)+2\sqrt{x-1}\sqrt{y-1})(z-1) = 1 \implies (x+y)(z-1) = 1 \implies z = \frac{1}{x+y} +1 = \frac{x^2+x-1}{x^2} \end{align*} Thus the solutions are the permutations of $(x, y, z) = \left(a, \frac{a}{a-1}, \frac{a^2+a-1}{a^2}\right)$ for $a > 1$. Would I lose one point if I wrote $a \ge 1$ instead of $a > 1$?
02.05.2013 01:07
tenniskidperson3 wrote:
x cannot be equal to 1
02.05.2013 01:07
That's what I did except I concluded that there were no solutions (Specifically, I set $s=b+c$ and $p=bc$.)
02.05.2013 01:08
Yay! I never knew that I was smart enough to get #6 on a USAJMO!
02.05.2013 01:09
hmm I got $(x, y, z)$ can be any permutation of $(\frac{1}{n}, \frac{1}{1-n}, -n^2+n+1)$ where $0<n<1$ though I might have screwed up entirely
02.05.2013 01:10
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=585&t=290260 darn. I saw this one before LOL.
02.05.2013 01:13
02.05.2013 01:14
Do you lose points if you have the right answer but have c=k^2+1 in the parameterization?
02.05.2013 01:14
02.05.2013 01:16
I wrote the triples as x, y, z without saying that $\sqrt{y-1}\sqrt{z-1} = 1$. In the last minute, I also added on for all y, z, since I forgot what my proof was about so I'll probably lose a lot of points on that. If you mention the product identity, though, you should be fine.
02.05.2013 01:18
How very original. Spent 3.5 hours trying to prove no solutions, then decided to randomly take quadratic formula, and it worked magically yay.
02.05.2013 01:28
Can someone please check my solution?
02.05.2013 01:34
All three people at our school who solved it are really bad at inequalities and decided to use calculus (presumably to minimize a function to find that it's minimized when x = 1+1/xy.) Then it falls from there.
02.05.2013 01:41
Wait you can just set the discriminant nonnegative and shift some stuff around.
02.05.2013 01:44
02.05.2013 01:46
james4l wrote: All three people at our school who solved it are really bad at inequalities and decided to use calculus (presumably to minimize a function to find that it's minimized when x = 1+1/xy.) Then it falls from there. I used calculus to find the minimums of some functions, but then I used AM-GM/ squares to actually prove that inequality.
02.05.2013 01:53
pi37 wrote: I used calculus to find the minimums of some functions. Will USAMO graders accept a proof using Calculus? For example using derivatives to show minimum or maximum points.
09.06.2020 03:23
The key is the following inequality along with its equality case. Lemma: For all nonnegative reals $a,b$, $\sqrt{a}+\sqrt{b} \le \sqrt{(a+1)(b+1)}$ with equality when $ab=1$. Solution: By AM-GM, $2\sqrt{ab}\le 1+ab$, with equality when $ab=1$. Add $a+b$ to both sides then take the square root of both sides to yield the desired. $\fbox{}$ WLOG $\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz}) = \sqrt{x+xyz}$. By this lemma, \[\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1}+\sqrt{yz}\le \sqrt{xyz+x}.\]Equality holds when $yz(x-1)=(y-1)(z-1)=1$. Then, reverse engineer to see that equality holds when \[\left(x,y,z\right) = \left(\frac{z^2+z-1}{z^2},\frac{z}{z-1},z\right).\]Note that cyclic variations thereof also constitute the equality case for some parameter $z$.
10.04.2021 09:09
The answer is $(x, y, z) = \left(\frac{t^2+t-1}{t^2}, t, \frac{t}{t-1} \right)$ and permutations for some $t > 1$. First, WLOG assume $x = \min (x, y, z)$. The LHS simplifies to $\sqrt{x+xyz}$. Next, we substitute $a=\sqrt{x-1}$, $b=\sqrt{y-1}$, and $c=\sqrt{z-1}$. Notice that $x=a^2+1$, $y=b^2+1$, and $z=c^2+1$, so the equation simplifies as \[ (a^2 + 1)((b^2 + 1)(c^2 + 1) + 1) = (a^2 + 1) + (a^2 + 1)(b^2 + 1)(c^2 + 1) = (a+b+c)^2.\]But notice that by Cauchy Schwarz, \[ (b^2 + 1)(c^2+1) = (b^2 + 1)(1+c^2) \ge (b+c)^2,\]and \[ (a^2 + 1)(1 + (b+c)^2) \ge (a+b+c)^2,\]but we must have equality, so $bc=1$ and $a(b+c)=1$. If we let $y=t$, then \[ z = c^2 + 1 = \left( \frac1{b^2} \right)^2 + 1 = \frac{t-1}+1 = \frac{t}{t-1}.\]We can also compute \[ x = a^2 + 1 = \left( \frac1{b+c} \right)^2 + 1 = \frac1{b^2 + c^2 + 2bc} + 1 = \frac1{b^2 + c^2 + 2} + 1 = \frac1{t + \frac{t}{t-1}} + 1 = \frac{t-1}{t^2} + 1 = \frac{t^2+t-1}{t^2},\]so we are done. $\blacksquare$
26.06.2021 19:08
Simple cauchy schwarz: WLOG $x \le y \le z$. First, notice that $$[(y-1)+1][1+(z-1)] \ge (\sqrt{y-1}+\sqrt{z-1})^2 \implies \sqrt{yz} \ge \sqrt{y-1}+\sqrt{z-1}.$$Also, we have $\sqrt{x+xyz} = \sqrt{x}\sqrt{1+yz}$, and $$[(x-1)+1][1+(yz)] \ge (\sqrt{x-1}+\sqrt{yz})^2 \implies \sqrt{x}\sqrt{yz+1} \ge \sqrt{x-1}+\sqrt{yz}.$$ Therefore, for equality to occur, we must have $\frac{y-1}{1} = \frac{1}{z-1}$, and $\frac{x-1}{1} = \frac{1}{yz}$. Now letting $c = y$ for some $c > 1$ gives the solutions $(\frac{c^2+c-1}{c^2}, c, \frac{c}{c-1})$ and its permutations.
21.08.2021 17:06
Nice and easy, solved with TheProblemIsSolved
31.08.2021 04:16
Let $x=a+1$, $y=b+1$ and $z=c+1$, and wlog $c=\min(a,b,c)$. Note that $a,b,c\geq 0$. Thus, we get the following equality, $$abc+ab+bc+ca+c+2=2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca},$$which is equivalent to $$c(a+1)(b+1)-\sqrt{c}(2\sqrt{b}+2\sqrt{c})+2-2\sqrt{ab}+ab=0.$$Solving the equality wrt $\sqrt{c}$, we get that \begin{align*} \sqrt{c}&=\frac{2\sqrt{a}+2\sqrt{b}\pm\sqrt{4(\sqrt{a}+\sqrt{b})^2-4(a+1)(b+1)(2+ab-2\sqrt{ab})}}{2(a+1)(b+1)}\\&=\frac{\sqrt{a}+\sqrt{b}\pm\sqrt{(\sqrt{a}+\sqrt{b})^2-(a+1)(b+1)(2+ab-2\sqrt{ab})}}{(a+1)(b+1)}. \end{align*}However, note that \begin{align*}(a+1)(b+1)(ab+2-2\sqrt{ab})&=(a+1)(b+1)((\sqrt{ab}-1)^2+1)\\&\geq (a+1)(b+1)\\&=a+b+ab+1\geq a+b+2\sqrt{ab}\\&=(\sqrt{a}+\sqrt{b})^2, \end{align*}where the equality holds iff $ab=1$. Indeed the equality must hold as otherwise the discriminant is negative. Thus, note that $$\sqrt{c}=\frac{a+\frac{1}{a}}{(1+a)(1+\frac{1}{a})}\implies c=\frac{a}{(a+1)^2}.$$We conclude that our solutions are in the form $(x,y,z)=(a+1,\frac{a+1}{a},\frac{a^2+3a+1}{(a+1)^2})$ for any $a>0$.
24.11.2021 19:02
WLOG $x\le y\le z$. So now \[\sqrt{x+xyz}=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\] We also have $\sqrt{x+xyz}=\sqrt{x}\cdot\sqrt{yz+1}$. By Cauchy-Schwarz, \[((x-1)+1)(1+yz))\ge (\sqrt{x-1}+\sqrt{yz})^2,\]so $\sqrt{x}\cdot\sqrt{yz+1}\ge \sqrt{x-1}+\sqrt{yz}$ since $x,y,z\ge 1$. Applying Cauchy-Schwarz again, we have \[(y-1)+1)(1+(z-1))\ge(\sqrt{y-1}+\sqrt{z-1})^2,\]so $\sqrt{yz}\ge \sqrt{y-1}+\sqrt{z-1}$. So we have \[\sqrt{x+xyz}\ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1},\]which implies equality must hold. Thus, \[\frac{\sqrt{x-1}}{1}=\frac{1}{\sqrt{yz}}\text{ and }\frac{\sqrt{y-1}}{1}=\frac{1}{\sqrt{z-1}}\] These simplify to $x-1=\frac{1}{yz}$ and $y-1=\frac{1}{z-1}$. This also implies that $z-1=\frac{1}{y-1}$. Now, we set $y=c$, to get $z=\frac{1}{c-1}+1=\frac{c}{c-1}$ and $x=\frac{c-1}{c^2}+1=\frac{c^2+c-1}{c^2}$. So our final answer is \[\boxed{\left(\frac{c^2+c-1}{c^2},c,\frac{c}{c-1}\right)}\]and its permutations for some $c>1$.
04.03.2022 19:44
Let $x=a^2+1,y=b^2+1,z=c^2+1$ and WLOG let $\min\{x,y,z\}=x$. Then the equation becomes \begin{align*} \sqrt{(a^2+1)((b^2+1)(c^2+1)+1)}&=a+b+c\\ (a^2+1)(b^2c^2+b^2+c^2+2)&=a^2+b^2+c^2+2ab+2bc+2ca\\ (b^2c^2+b^2+c^2+1)a^2-2(b+c)a+(b^2c^2-2bc+2)&=0\\ (b^2+1)(c^2+1)a^2-2(b+c)a+(1+(bc-1)^2)&=0. \end{align*}By the quadratic formula (discriminant) this has a real solution in $a$ only if $$4(b+c)^2 \geq 4(b^2+1)(c^2+1)(1+(bc-1)^2) \implies (b+c)^2 \geq (b^2+1)(c^2+1) \implies 0 \geq (bc-1)^2,$$which is only true for $bc=1$, where the discriminant is zero. Then letting $b=k$ for $k \geq 0$, we have $c=\tfrac{1}{k}$ and $$a=\frac{2(b+c)}{(b^2+1)(c^2+1)}=\frac{k+\frac{1}{k}}{k^2+2+\frac{1}{k^2}}=\frac{1}{k+\frac{1}{k}},$$which is always the least among $b,c$, so we in fact have $\min\{x,y,z\}=x$. From here we extract $$(x,y,z)=\left(1+\frac{1}{(k+\tfrac{1}{k})^2},1+k^2,1+\frac{1}{k^2}\right)~\forall k \geq 0$$and permutations, which can be checked to work. $\blacksquare$
04.04.2022 21:19
Claim: We have $\sqrt{x-1}+\sqrt{y-1}\leq\sqrt{xy}$ with equality if and only if $x+y=xy$. Proof: Squaring, it suffices to show that $x+y-2+2\sqrt{(x-1)(y-1)}\leq xy$, so if $a=xy-x-y$, then $2\sqrt{a+1}\leq a+2$, which follows from AM-GM since $\frac{(a+1)+1}2\geq\sqrt{a+1}$, with equality if and only if $a=0$, or $x+y=xy$. Therefore, if we assume without loss of generality $x\geq y\geq z$, then we have \begin{align*} \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}&\leq\sqrt{(xy+1)-1}+\sqrt{z-1}\\ &\leq\sqrt{xyz+z}\\ &=\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz}). \end{align*}The only time both sides can be equal is when $x+y=xy$ and $(xy+1)z=xy+z+1$. This is equivalent to $(x-1)(y-1)=1$ and $z=1+\frac1{xy}$. Therefore, we have $x=1+m$, $y=1+\frac1m$, and $z=1+\frac m{(m+1)^2}$, so the solutions are $\boxed{\left(1+m,1+\frac1m,1+\frac m{(m+1)^2}\right)}$ and permutations.
04.07.2022 09:54
By CS note that for any $r,s,t:$ $$(1+(r+s)^2)(1+t^2)\geq (r+s+t)^2$$Set $(x,y,z)=(1+r^2,1+s^2,1+t^2)$ so $$(r+s)^2\geq (1+r^2)+(1+s^2)\iff (rs-1)^2\leq 0$$Therefore $rs=1$ and $t=\tfrac{1}{r+s}$ Our solution is $$\{x,y,z\}=\left \{1+r^2,1+\frac{1}{r^2}, 1+\left(\frac{r}{r^2+1}\right )^2\right \}$$which works.
06.08.2022 02:46
03.05.2023 23:10
Let $(x,y,z) = (a^2 + 1, b^2 + 1, c^2+1)$ with $a,b,c\ge 0$ and WLOG $a = \min(a,b,c)$. Then, we have \[ \sqrt{a^2 + 1+ (a^2 + 1)(b^2+ 1)(c^2+1)} = a + b+ c\]\[ \implies (a^2 + 1)((b^2 + 1)(c^2 + 1) + 1) = (a +b + c)^2.\]Note that $(b^2 + 1)(c^2 + 1) = ((bc)^2 + 1) + b^2 + c^2\ge (2bc) + b^2 + c^2 = (b+ c)^2$, with equality when $bc = 1$. Thus, we have \[ (a + b +c)^2\ge (a^2 + 1)(1 + (b+c)^2).\]However, by Cauchy, we have $(a^2 + 1)(1 + (b+ c)^2)\ge (a + b + c)^2$, so equality in both the above inequality as well as the inequality $(b^2 + 1)(c^2 +1)\ge (b + c)^2$ must hold. That is, $(a, 1)$ and $(1, b+c)$ are proportional and $bc = 1$. This is easy to solve, and we get \[(x,y,z) = \left(1 + \left(\frac{1}{b + \frac{1}{b}}\right)^2, 1 + b^2, 1 + \frac{1}{b^2}\right),\]as well as permutations, for any $b\in\mathbb{R}_{>0}$.
26.05.2024 22:48
Our claimed solution set, which we can confirm to work, as the permutations of \[\boxed{\left(\frac{y^2+y-1}{y^2}, y, \frac{y}{y-1}\right), \quad y > 1}.\] Now assume WLOG $x = \min(x,y,z)$. Our equation is then \begin{align*} \sqrt{x+xyz} &= \sqrt{x-1} + (\sqrt{y-1} + \sqrt{z-1}) \\ &\leq \sqrt{x-1} + \sqrt{(1+(y-1)) \cdot ((z-1)+1)} \\ &= \sqrt{x-1} + \sqrt{(yz+1)-1} \\ &\leq \sqrt{(1+(x-1)) \cdot ((yz+1)+1)} \\ &= \sqrt{x+xyz}, \end{align*} where we have the inequalities by Cauchy. We must have the two equality cases \[(x-1)(y-1) = (x-1)(yz+1) = 1,\] from which we get the solutions above. $\blacksquare$
24.08.2024 20:30
Nice and easy! Let $(x,y,z)=\left (a^2+1,b^2+1,c^2+1\right )$ and suppose wlog that $0\leq a\leq b\leq c$. Then $$\sqrt{(a^2+1)+(a^2+1)(b^2+1)(c^2+1)}=a+b+c \ \Rightarrow \ a^2\left (b^2c^2+b^2+c^2+1\right )-a(2b+2c)+\left (b^2c^2-2bc+2\right )=0.$$The discriminant is $$4(b+c)^2-4\left (b^2c^2-2bc+2\right )\left (b^2c^2+b^2+c^2+1\right )=-4(bc-1)^2\left(c^2b^2+c^2+b^2+2\right )\leq 0$$and so it follows that $c=1/b$. And we also find $a=\frac{b}{b^2+1}$. Conversely it's easy to check that $$(x,y,z)=\left(\frac{t^2}{(t^2+1)^2}+1,t^2+1,\frac 1{t^2}+1\right)$$works.
07.12.2024 19:19
Pretty cool and conceptual kind of problem. We claim that the answer is $(x,y,z)=\left(\frac{a}{(a+1)^2}+1,a+1,\frac{1}{a}+1\right)$ for all positive real numbers $a$. It is (not) easy to see that these triples so. Now we shall show that they are the only ones. For this, we first show the following inequality. Claim : For all positive real numbers $a$ and $b$, \[\sqrt{a} + \sqrt{b} \le \sqrt{(a+1)(b+1)}\]with equality if and only if $ab=1$. Proof : Note that, \begin{align*} (\sqrt{ab}-1)^2 & \ge 0\\ ab -2\sqrt{ab} +1 & \ge 0\\ a + 2 \sqrt{ab} + b & \le ab + a + b + 1\\ (\sqrt{a}+\sqrt{b})^2 & \le (a+1)(b+1)\\ \sqrt{a} + \sqrt{b} & \le \sqrt{(a+1)(b+1)} \end{align*}as we set out to show, with equality clearly if and only if $ab=1$. Now, say the minimum among $x,y,z$ is $x$. Then applying the above lemma twice we note, \[\sqrt{x-1} + \sqrt{y-1} + \sqrt{z-1} \ge \sqrt{x-1} + \sqrt{yz} \ge \sqrt{x(yz+1)}=\sqrt{x+xyz}\]with equality if and only if equality holds in each inequality. Thus, we are left to solve the system of equations, \[(y-1)(z-1)=1 \text{ and } (x-1)yz=1\]First, note that the first equation rewrites to, \[yz=y+z\]Let $y=a+1$ and $z=b+1$ for non-negative reals $a$ and $b$. Then, \[ab+a+b+1=(a+1)(b+1)=(a+1)+(b+1)=a+b+2\]which simplifies to $ab=1$. Thus, we have that \[y=a+1 \text{ and } z= \frac{1}{a}+1\]Substituting these results into the second equation we obtain that, \[x=\frac{a}{(a+1)^2}+1\]and indeed all solutions take the desired forms.
03.01.2025 05:57
Video solution