Difference between revisions of "1984 IMO Problems/Problem 1"
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<math>0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}</math> | <math>0 \leq xy+yz+zx-2xyz \leq \frac{7}{27}</math> | ||
− | ==Solution== | + | ==Solution 1== |
Note that this inequality is symmetric with x,y and z. | Note that this inequality is symmetric with x,y and z. | ||
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To prove <math>xy+yz+zx-2xyz \leq \frac{7}{27}</math>, suppose <math>x \leq y \leq z</math>. Note that <math>x \leq y \leq \frac{1}{2}</math> since at most one of x,y,z is <math>\frac{1}{2}</math>. Suppose not all of them equals <math>\frac{1}{3}</math>-otherwise, we would be done. This implies <math>x \leq \frac{1}{3}</math> and <math>z \geq \frac{1}{3}</math>. Thus, define <cmath>x' =\frac{1}{3}</cmath>, <cmath>y'=y</cmath> <cmath>z'=x+y-\frac{1}{3}</cmath> <cmath>\epsilon = \frac{1}{3}-x</cmath> Then, <math>x'=x+\epsilon</math>, <math>z'=z-\epsilon</math>, and <math>x'+y'+z'=1</math>. After some simplification, <cmath>x'y'+y'z'+z'x'-2x'y'z'=xy+yz+zx-2xyz+(1-2y)(z-x-\epsilon)>xy+yz+zx-2xyz</cmath> since <math>1-2y>0</math> and <math>z-x-\epsilon=z-\frac{1}{3}>0</math>. If we repeat the process, defining <cmath>x'' =x'=\frac{1}{3}</cmath> <cmath>y''=\frac{1}{3}</cmath> <cmath>z''=z'+y'-\frac{1}{3}=\frac{1}{3}</cmath> after similar reasoning, we see that <cmath>xy+yz+zx-2xyz\leq x'y'+y'z'+z'x'-2x'y'z' \leq x''y''+y''z''+z''x''-2x''y''z''=\frac{7}{27}</cmath>. | To prove <math>xy+yz+zx-2xyz \leq \frac{7}{27}</math>, suppose <math>x \leq y \leq z</math>. Note that <math>x \leq y \leq \frac{1}{2}</math> since at most one of x,y,z is <math>\frac{1}{2}</math>. Suppose not all of them equals <math>\frac{1}{3}</math>-otherwise, we would be done. This implies <math>x \leq \frac{1}{3}</math> and <math>z \geq \frac{1}{3}</math>. Thus, define <cmath>x' =\frac{1}{3}</cmath>, <cmath>y'=y</cmath> <cmath>z'=x+y-\frac{1}{3}</cmath> <cmath>\epsilon = \frac{1}{3}-x</cmath> Then, <math>x'=x+\epsilon</math>, <math>z'=z-\epsilon</math>, and <math>x'+y'+z'=1</math>. After some simplification, <cmath>x'y'+y'z'+z'x'-2x'y'z'=xy+yz+zx-2xyz+(1-2y)(z-x-\epsilon)>xy+yz+zx-2xyz</cmath> since <math>1-2y>0</math> and <math>z-x-\epsilon=z-\frac{1}{3}>0</math>. If we repeat the process, defining <cmath>x'' =x'=\frac{1}{3}</cmath> <cmath>y''=\frac{1}{3}</cmath> <cmath>z''=z'+y'-\frac{1}{3}=\frac{1}{3}</cmath> after similar reasoning, we see that <cmath>xy+yz+zx-2xyz\leq x'y'+y'z'+z'x'-2x'y'z' \leq x''y''+y''z''+z''x''-2x''y''z''=\frac{7}{27}</cmath>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | By the method of Lagrangian multipliers. Let <math>f(x,y,z) = xy + yz + zx - 2xyz</math> and <math>\phi(x,y,z) = x+y+z-1</math>. We will find the local maxima/minima of <math>f</math> over <math>S = \{(x,y,z) \in \mathbb{R}^2 : 0 \leq x,y,z \leq 1\}</math> subject to <math>\phi = 0</math>. | ||
+ | |||
+ | Let <math>S' = S \cap \{(x,y,z) \in \mathbb{R}^2 : \phi(x,y,z) = 0\}</math>. <math>S'</math> is bounded and the intersection of two closed sets, hence compact. So every sequence in <math>S'</math> has a convergent subsequence. Hence if <math>(a_k)_k \subset S'</math> is such that <math>f(a_k)</math> converges to the infimum or supremum of <math>f</math> in <math>S'</math>, there will be a convergent subsequence <math>a_{k_j} \rightarrow a_\infty \in S'</math> and <math>f(a_\infty) = \lim_{j \rightarrow \infty}f(a_{k_j})</math> by the continuity of <math>f</math>. Hence the infimum/supremum of <math>f</math> over <math>S'</math> will also be a local minima/maxima of <math>f</math> over <math>S'</math>. | ||
+ | |||
+ | We must solve <math>\nabla f - \lambda \nabla \phi = 0</math>. This is equivalent to | ||
+ | \begin{align} | ||
+ | x+y - 2xy &= -2uv + 1/2 = \lambda \\ | ||
+ | y+z - 2yz &= -2vw + 1/2 = \lambda \\ | ||
+ | z+x - 2zx &= -2wu + 1/2 = \lambda | ||
+ | \end{align} | ||
+ | where <math>u = 1/2 - x, v = 1/2 - y, w = 1/2 - z</math>. If <math>\lambda = 1/2</math> then <math>uv = vw = wu = 0</math>. WLOG <math>u= 0</math> and we have <math>vw = 0</math>. | ||
+ | Then WLOG <math>v = 0</math>. These imply <math>x = y = 1/2</math>. Then <math>z = 0</math> since <math>x+y+z = 1</math>. We get <math>f(1/2,1/2,0) = 1/4 < 7/27</math>. | ||
+ | |||
+ | If <math>\lambda \neq 1/2</math> then letting <math>\lambda' = 1/4 - \lambda/2 \neq 0</math> one gets <math>uv = vw = wu = \lambda'</math> which imply <math>u=v=w</math> since <math>u,v,w \neq 0</math>. This implies <math>x=y=z=1/3</math> since <math>x+y+z=1</math>. And <math>f(1/3,1/3,1/3) = 7/27</math>. | ||
+ | |||
+ | Now to check the boundary of <math>0 \leq x,y,z \leq 1</math>. WLOG we must consider the cases <math>z=1</math> and <math>z=0</math>. If <math>z=1</math> then <math>x=y=0</math> by <math>x+y+z=1</math> so <math>f(x,y,z) = 0</math>. If <math>z = 0</math> substituting <math>y=1-x</math> in <math>f(x,y,0)</math> yields | ||
+ | <cmath> | ||
+ | f(x,1-x,0) = x(1-x) = -(1/2-x)^2+1/4. | ||
+ | </cmath> | ||
+ | which is between <math>0</math> and <math>1/4<7/27</math> since <math>0 \leq x \leq 1</math>. Considering all the values found we find <math>0 \leq f(x,y,z) \leq 7/27</math>. | ||
+ | |||
+ | ~not_detriti | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 12:32, 10 July 2025
Problem
Let ,
,
be nonnegative real numbers with
. Show that
Solution 1
Note that this inequality is symmetric with x,y and z.
To prove note that
implies that at most one of
,
, or
is greater than
. Suppose
, WLOG. Then,
since
, implying all terms are positive.
To prove , suppose
. Note that
since at most one of x,y,z is
. Suppose not all of them equals
-otherwise, we would be done. This implies
and
. Thus, define
,
Then,
,
, and
. After some simplification,
since
and
. If we repeat the process, defining
after similar reasoning, we see that
.
Solution 2
By the method of Lagrangian multipliers. Let and
. We will find the local maxima/minima of
over
subject to
.
Let .
is bounded and the intersection of two closed sets, hence compact. So every sequence in
has a convergent subsequence. Hence if
is such that
converges to the infimum or supremum of
in
, there will be a convergent subsequence
and
by the continuity of
. Hence the infimum/supremum of
over
will also be a local minima/maxima of
over
.
We must solve . This is equivalent to
\begin{align}
x+y - 2xy &= -2uv + 1/2 = \lambda \\
y+z - 2yz &= -2vw + 1/2 = \lambda \\
z+x - 2zx &= -2wu + 1/2 = \lambda
\end{align}
where
. If
then
. WLOG
and we have
.
Then WLOG
. These imply
. Then
since
. We get
.
If then letting
one gets
which imply
since
. This implies
since
. And
.
Now to check the boundary of . WLOG we must consider the cases
and
. If
then
by
so
. If
substituting
in
yields
which is between
and
since
. Considering all the values found we find
.
~not_detriti
Video Solution
Video Solution
See Also
1984 IMO (Problems) • Resources | ||
Preceded by First Question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |