2011 IMO Problems/Problem 3: Difference between revisions
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==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== | ==See Also== | ||
{{IMO box|year=2011|num-b=2|num-a=4}} | |||
[[Category:Olympiad Algebra Problems]] | |||
[[Category:Functional Equation Problems]] | |||
Latest revision as of 00:20, 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 | ||