# 2013 USAMO Problems/Problem 4

Find all real numbers $x,y,z\geq 1$ satisfying $$\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.$$

## Solution (Cauchy or AM-GM)

The key Lemma is: $$\sqrt{a-1}+\sqrt{b-1} \le \sqrt{ab}$$ for all $a,b \ge 1$. Equality holds when $(a-1)(b-1)=1$.

This is proven easily. $$\sqrt{a-1}+\sqrt{b-1} = \sqrt{a-1}\sqrt{1}+\sqrt{1}\sqrt{b-1} \le \sqrt{(a-1+1)(b-1+1)} = \sqrt{ab}$$ by Cauchy. Equality then holds when $a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1$.

Now assume that $x = \min(x,y,z)$. Now note that, by the Lemma,

$$\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}$$. So equality must hold. So $(y-1)(z-1) = 1$ and $(x-1)(yz) = 1$. If we let $z = c$, then we can easily compute that $y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}$. Now it remains to check that $x \le y, z$.

But by easy computations, $x = \frac{c^2+c-1}{c^2} \le c = z \Longleftrightarrow (c^2-1)(c-1) \ge 0$, which is obvious. Also $x = \frac{c^2+c-1}{c^2} \le \frac{c}{c-1} = y \Longleftrightarrow 2c \ge 1$, which is obvious, since $c \ge 1$.

So all solutions are of the form $\boxed{\left(\frac{c^2+c-1}{c^2}, \frac{c}{c-1}, c\right)}$, and all permutations for $c > 1$.

Remark: An alternative proof of the key Lemma is the following: By AM-GM, $$(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}$$ $$ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}$$. Now taking the square root of both sides gives the desired. Equality holds when $(a-1)(b-1) = 1$.

## Solution with Thought Process

Without loss of generality, let $1 \le x \le y \le z$. Then $\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}$.

Suppose x = y = z. Then $\sqrt{x + x^3} = 3\sqrt{x-1}$, so $x + x^3 = 9x - 9$. It is easily verified that $x^3 - 8x + 9 = 0$ has no solution in positive numbers greater than 1. Thus, $\sqrt{x + xyz} \ge \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}$ for x = y = z. We suspect if the inequality always holds.

Let x = 1. Then we have $\sqrt{1 + yz} \ge \sqrt{y-1} + \sqrt{z-1}$, which simplifies to $$1 + yz \ge y + z - 2 + 2\sqrt{(y-1)(z-1)}$$ and hence $$yz - y - z + 3 \ge 2\sqrt{(y-1)(z-1)}$$ Let us try a few examples: if y = z = 2, we have $3 > 2$; if y = z, we have $y^2 - 2y + 3 \ge 2(y-1)$, which reduces to $y^2 - 4y + 5 \ge 0$. The discriminant (16 - 20) is negative, so in fact the inequality is strict. Now notice that yz - y - z + 3 = (y-1)(z-1) + 2. Now we see we can let $u = \sqrt{(y-1)(z-1)}$! Thus, $$u^2 - 2u + 2 = (u-1)^2 + 1 > 0$$ and the claim holds for x = 1.

If x > 1, we see the $\sqrt{x - 1}$ will provide a huge obstacle when squaring. But, using the identity $(x+y+z)^2 = x^2 + y^2 + z^2 + xy + yz + xz$: $$x + xyz \ge x - 1 + y - 1 + z - 1 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(y-1)}$$ which leads to $$xyz \ge y + z - 3 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(z-1)}$$ Again, we experiment. If x = 2, y = 3, and z = 3, then $18 > 7 + 4\sqrt{6}$.

Now, we see the finish: setting $u = \sqrt{x-1}$ gives $x = u^2 + 1$. We can solve a quadratic in u! Because this problem is a #6, the crown jewel of USAJMO problems, we do not hesitate in computing the messy computations: $$u^2(yz) - u(2\sqrt{y-1} + 2\sqrt{z-1}) + (3 + yz - y - z - 2\sqrt{(y-1)(z-1)}) \ge 0$$

Because the coefficient of $u^2$ is positive, all we need to do is to verify that the discriminant is nonpositive: $$b^2 - 4ac = 4(y-1) + 4(z-1) - 8\sqrt{(y-1)(z-1)} - yz(12 + 4yz - 4y - 4z - 8\sqrt{(y-1)(z-1)})$$

Let us try a few examples. If y = z, then the discriminant D = $8(y-1) - 8(y-1) - yz(12 + 4y^2 - 8y - 8(y-1)) = -yz(4y^2 - 16y + 20) = -4yz(y^2 - 4y + 5) < 0$.

We are almost done, but we need to find the correct argument. (How frustrating!) Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are no real solutions to the original equation in the hypothesis.

--Thinking Process by suli

## Solution 2

WLOG, assume that $x = \min(x,y,z)$. Let $a=\sqrt{x-1},$ $b=\sqrt{y-1}$ and $c=\sqrt{z-1}$. Then $x=a^2+1$, $y=b^2+1$ and $z=c^2+1$. The equation becomes $$(a^2+1)+(a^2+1)(b^2+1)(c^2+1)=(a+b+c)^2.$$ Rearranging the terms, we have $$(1+a^2)(bc-1)^2+[a(b+c)-1]^2=0.$$ Therefore $bc=1$ and $a(b+c)=1.$ Express $a$ and $b$ in terms of $c$, we have $a=\frac{c}{c^2+1}$ and $b=\frac{1}{c}.$ Easy to check that $a$ is the smallest among $a$, $b$ and $c.$ Then $x=\frac{c^4+3c^2+1}{(c^2+1)^2}$, $y=\frac{c^2+1}{c^2}$ and $z=c^2+1.$ Let $c^2=t$, we have the solutions for $(x,y,z)$ as follows: $(\frac{t^2+3t+1}{(t+1)^2}, \frac{t+1}{t}, t+1)$ and permutations for all $t>0.$

--J.Z.