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 ![\[1 + yz \ge y + z - 2 + 2\sqrt{(y-1)(z-1)}\]](http://latex.artofproblemsolving.com/1/5/e/15ea895126a2309ec5d0795354dac079c7cc25f3.png)
and hence ![\[yz - y - z + 3 \ge 2\sqrt{(y-1)(z-1)}\]](http://latex.artofproblemsolving.com/3/9/e/39eb443cc2a760eb03d2d95bfabdb1123da6759f.png)
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, ![\[u^2 - 2u + 2 = (u-1)^2 + 1 > 0\]](http://latex.artofproblemsolving.com/3/4/b/34bc0d41229e722d449cac8de681088b1eb6691d.png)
and the claim holds for x = 1.
If x > 1, we see the 
will provide a huge obstacle when squaring. But, using the identity 
:
![\[x + xyz \ge x - 1 + y - 1 + z - 1 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(y-1)}\]](http://latex.artofproblemsolving.com/9/c/a/9ca4391efb211202afb93d416503511aee16aeca.png)
which leads to
![\[xyz \ge y + z - 3 + 2\sqrt{(x-1)(y-1)} + 2\sqrt{(y-1)(z-1)} + 2\sqrt{(x-1)(z-1)}\]](http://latex.artofproblemsolving.com/2/0/3/203dc435bb009fd94b759d273476a2c6b3a0b1da.png)
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:
![\[u^2(yz) - u(2\sqrt{y-1} + 2\sqrt{z-1}) + (3 + yz - y - z - 2\sqrt{(y-1)(z-1)}) \ge 0\]](http://latex.artofproblemsolving.com/b/b/a/bba13029b17812c8252f321e828e1905af983f89.png)
Because the coefficient of 
is greater than 0, all we need to do is to verify that the discriminant is nonpositive:
![\[b^2 - 4ac = 4(y-1) + 4(z-1) - 8\sqrt{(y-1)(z-1)} - yz(12 - 4yz + 4y + 4z + 8\sqrt{(y-1)(z-1)})\]](http://latex.artofproblemsolving.com/6/8/8/68878d095512153524916434ebe9b62f3d545930.png)
<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
