Difference between revisions of "2011 IMO Problems/Problem 3"
m |
|||
(3 intermediate revisions by 3 users not shown) | |||
Line 2: | Line 2: | ||
==Solution== | ==Solution== | ||
− | {{ | + | Let <math>P(x,y)</math> be the given assertion. |
+ | Comparing <math>P(x,f(y)-x)</math> and <math>P(y,f(x)-y)</math> yields, <cmath>xf(x)+yf(y)\leq 2f(x)f(y).</cmath> | ||
+ | <math>y\mapsto 2f(x)\implies xf(x)\leq 0. \qquad (*)</math> | ||
+ | ------------------------ | ||
+ | <math>\textbf{Claim: }f(k)\leq 0~~\forall k.</math> | ||
+ | |||
+ | <math>Proof.</math> Suppose <math>\exists k:f(k)>0,</math> then <cmath>f(k+y)\leq yf(k)+f(f(k)).</cmath> | ||
+ | Now <math>y\to -\infty</math> implies that <math>\lim_{x\to -\infty} f(x)=-\infty.</math> | ||
+ | <math>P(x,z-x)\implies f(z)\leq (z-x)f(x)+f(f(x)).</math> | ||
+ | |||
+ | Then <math>x\to -\infty,</math> yields a contradiction. <math>\blacksquare</math> | ||
+ | -------------------------- | ||
+ | From <math>(*)</math> we get <math>f(x)=0,\forall x<0.</math> | ||
+ | <math>P(0,f(0))\implies f(0)\geq 0,</math> thus we get <math>f(0)=0,</math> as desired. <math>\square</math> | ||
+ | |||
+ | ~ZETA_in_olympiad | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2011|num-b=2|num-a=4}} | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] |
Latest revision as of 01:20, 19 November 2023
Let be a real-valued function defined on the set of real numbers that satisfies
for all real numbers
and
. Prove that
for all
.
Solution
Let be the given assertion.
Comparing
and
yields,
Suppose
then
Now
implies that
Then yields a contradiction.
From we get
thus we get
as desired.
~ZETA_in_olympiad
See Also
2011 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |