DARN FORGOT TO CONSIDER EQUALITY CASE

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 $a > 1$?

tenniskidperson3 wrote:

x cannot be equal to 1

That's what I did except I concluded that there were no solutions (Specifically, I set $s=b+c$ and $p=bc$.)

Yay! I never knew that I was smart enough to get #6 on a USAJMO!

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

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=585&t=290260 darn. I saw this one before LOL.

Do you lose points if you have the right answer but have c=k^2+1 in the parameterization?

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.

How very original. Spent 3.5 hours trying to prove no solutions, then decided to randomly take quadratic formula, and it worked magically yay.

Can someone please check my solution?

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.

Wait you can just set the discriminant nonnegative and shift some stuff around.

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.

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.

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 $z$.

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. $\blacksquare$

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.

Nice and easy, solved with TheProblemIsSolved

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 $a>0$.

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 $c>1$.

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. $\blacksquare$

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.

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.

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 $b\in\mathbb{R}_{>0}$.

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. $\blacksquare$

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.

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.

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