Difference between revisions of "2013 USAJMO Problems/Problem 6"
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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> | <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> | ||
− | Because u > 0, all we need to do is to verify that the discriminant is nonpositive: | + | Because the coefficient of <math>u^2</math> is greater than 0, all we need to do is to verify that the discriminant is nonpositive: |
<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> | <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> | ||
<prove that discriminant is negative> | <prove that discriminant is negative> | ||
Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are <math>\boxed{no real solutions}</math> to the original equation in the hypothesis. | Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are <math>\boxed{no real solutions}</math> to the original equation in the hypothesis. | ||
--Thinking Process by suli | --Thinking Process by suli |
Revision as of 11:43, 14 April 2014
Solution with Thought Process
Without loss of generality, let . Then
.
Suppose x = y = z. Then , so
. It is easily verified that
has no solution in positive numbers greater than 1. Thus,
for x = y = z. We suspect if the inequality always holds.
Let x = 1. Then we have , which simplifies to
and hence
Let us try a few examples: if y = z = 2, we have
; if y = z, we have
, which reduces to
. 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
! Thus,
and the claim holds for x = 1.
If x > 1, we see the will provide a huge obstacle when squaring. But, using the identity
:
which leads to
Again, we experiment. If x = 2, y = 3, and z = 3, then
.
Now, we see the finish: setting gives
. 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:
Because the coefficient of is greater than 0, all we need to do is to verify that the discriminant is nonpositive:
<prove that discriminant is negative>
Success! The discriminant is negative. Thus, we can replace our claim with a strict one, and there are
to the original equation in the hypothesis.
--Thinking Process by suli