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Difference between revisions of "2013 USAMO Problems/Problem 4"
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== Problem ==
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Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath>
 
Find all real numbers <math>x,y,z\geq 1</math> satisfying <cmath>\min(\sqrt{x+xyz},\sqrt{y+xyz},\sqrt{z+xyz})=\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}.</cmath>
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== Solution  (Cauchy or AM-GM) ==
 
== Solution  (Cauchy or AM-GM) ==
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This is proven easily.
 
This is proven easily.
 
<cmath>\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}</cmath> by Cauchy.
 
<cmath>\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}</cmath> by Cauchy.
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Equality then holds when <math>a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1</math>.
 
Equality then holds when <math>a-1 =\frac{1}{b-1} \implies (a-1)(b-1) = 1</math>.
  
 
Now assume that <math>x = \min(x,y,z)</math>. Now note that, by the Lemma,
 
Now assume that <math>x = \min(x,y,z)</math>. Now note that, by the Lemma,
  
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<cmath>\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}</cmath>. So equality must hold.
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<cmath>\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1} \le \sqrt{x-1} + \sqrt{yz} \le \sqrt{x(yz+1)} = \sqrt{xyz+x}</cmath>. So equality must hold in order for the condition in the problem statement to be met.
 
So <math>(y-1)(z-1) = 1</math> and <math>(x-1)(yz) = 1</math>. If we let <math>z = c</math>, then we can easily compute that <math>y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}</math>.
 
So <math>(y-1)(z-1) = 1</math> and <math>(x-1)(yz) = 1</math>. If we let <math>z = c</math>, then we can easily compute that <math>y = \frac{c}{c-1}, x = \frac{c^2+c-1}{c^2}</math>.
 
Now it remains to check that <math>x \le y, z</math>.
 
Now it remains to check that <math>x \le y, z</math>.
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By AM-GM, <cmath>(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}</cmath>
 
By AM-GM, <cmath>(ab-a-b+1)+1 = (a-1)(b-1) + 1 \ge 2\sqrt{(a-1)(b-1)}</cmath>
 
<cmath>ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}</cmath>. Now taking the square root of both sides gives the desired. Equality holds when <math>(a-1)(b-1) = 1</math>.
 
<cmath>ab\ge (a-1)+(b-1)+2\sqrt{(a-1)(b-1)}</cmath>. Now taking the square root of both sides gives the desired. Equality holds when <math>(a-1)(b-1) = 1</math>.
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==Solution with Thought Process==
 
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Without loss of generality, let <math>1 \le x \le y \le z</math>. Then <math>\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math>.
 
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Suppose x = y = z. Then <math>\sqrt{x + x^3} = 3\sqrt{x-1}</math>, so <math>x + x^3 = 9x - 9</math>. It is easily verified that <math>x^3 - 8x + 9 = 0</math> has no solution in positive numbers greater than 1. Thus, <math>\sqrt{x + xyz} \ge \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}</math> for x = y = z. We suspect if the inequality always holds.
 
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Let x = 1. Then we have <math>\sqrt{1 + yz} \ge \sqrt{y-1} + \sqrt{z-1}</math>, which simplifies to <cmath>1 + yz \ge y + z - 2 + 2\sqrt{(y-1)(z-1)}</cmath> and hence <cmath>yz - y - z + 3 \ge 2\sqrt{(y-1)(z-1)}</cmath> Let us try a few examples: if y = z = 2, we have <math>3 > 2</math>; if y = z, we have <math>y^2 - 2y + 3 \ge 2(y-1)</math>, which reduces to <math>y^2 - 4y + 5 \ge 0</math>. 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 <math>u = \sqrt{(y-1)(z-1)}</math>! Thus, <cmath>u^2 - 2u + 2 = (u-1)^2 + 1 > 0</cmath> and the claim holds for x = 1.
 
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If x > 1, we see the <math>\sqrt{x - 1}</math> will provide a huge obstacle when squaring. But, using the identity <math>(x+y+z)^2 = x^2 + y^2 + z^2 + xy + yz + xz</math>:
 
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<cmath>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)}</cmath>
 
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which leads to
 
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<cmath>xyz \ge y + z - 3 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(z-1)}</cmath>
 
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Again, we experiment. If x = 2, y = 3, and z = 3, then <math>18 > 7 + 4\sqrt{6}</math>.
 
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Now, we see the finish: setting <math>u = \sqrt{x-1}</math> gives <math>x = u^2 + 1</math>. 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:
 
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<cmath>u^2(yz) - u(2\sqrt{y-1} + 2\sqrt{z-1}) + (3 + yz - y - z - 2\sqrt{(y-1)(z-1)}) \ge 0</cmath>
 
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Because the coefficient of <math>u^2</math> is positive, all we need to do is to verify that the discriminant is nonpositive:
 
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<cmath>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)})</cmath>
 
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Let us try a few examples. If y = z, then the discriminant D = <math>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</math>.
 
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We are almost done, but we need to find the correct argument. (How frustrating!)
 
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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.
 
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--Thinking Process by suli
 
  
 
== Solution 2  ==
 
== Solution 2  ==

Latest revision as of 13:36, 28 October 2024

Problem

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 in order for the condition in the problem statement to be met. 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 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.

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