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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>. |
− | ==(Incorrect) 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
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| == Solution 2 == | | == Solution 2 == |
| WLOG, assume that <math>x = \min(x,y,z)</math>. Let <math>a=\sqrt{x-1},</math> <math>b=\sqrt{y-1}</math> and <math>c=\sqrt{z-1}</math>. Then <math>x=a^2+1</math>, <math>y=b^2+1</math> and <math>z=c^2+1</math>. The equation becomes | | WLOG, assume that <math>x = \min(x,y,z)</math>. Let <math>a=\sqrt{x-1},</math> <math>b=\sqrt{y-1}</math> and <math>c=\sqrt{z-1}</math>. Then <math>x=a^2+1</math>, <math>y=b^2+1</math> and <math>z=c^2+1</math>. The equation becomes |
Revision as of 19:02, 13 June 2020
Find all real numbers satisfying
Solution (Cauchy or AM-GM)
The key Lemma is:
for all . Equality holds when .
This is proven easily.
by Cauchy.
Equality then holds when .
Now assume that . Now note that, by the Lemma,
. So equality must hold.
So and . If we let , then we can easily compute that .
Now it remains to check that .
But by easy computations, , which is obvious.
Also , which is obvious, since .
So all solutions are of the form , and all permutations for .
Remark: An alternative proof of the key Lemma is the following:
By AM-GM,
. Now taking the square root of both sides gives the desired. Equality holds when .
Solution 2
WLOG, assume that . Let and . Then , and . The equation becomes
Rearranging the terms, we have
Therefore and Express and in terms of , we have and Easy to check that is the smallest among , and Then , and
Let , we have the solutions for as follows:
and permutations for all
--J.Z.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.