2011 IMO Problems/Problem 3: Difference between revisions

Latest revision as of 00:20, 19 November 2023

Let $f: \mathbb R \to \mathbb R$ be a real-valued function defined on the set of real numbers that satisfies $f(x + y) \le yf(x) + f(f(x))$ for all real numbers $x$ and $y$ . Prove that $f(x) = 0$ for all $x \le 0$ .

Solution

Let $P(x,y)$ be the given assertion. Comparing $P(x,f(y)-x)$ and $P(y,f(x)-y)$ yields, $xf(x)+yf(y)\leq 2f(x)f(y).$ $y\mapsto 2f(x)\implies xf(x)\leq 0. \qquad (*)$

$\textbf{Claim: }f(k)\leq 0~~\forall k.$

$Proof.$ Suppose $\exists k:f(k)>0,$ then $f(k+y)\leq yf(k)+f(f(k)).$ Now $y\to -\infty$ implies that $\lim_{x\to -\infty} f(x)=-\infty.$ $P(x,z-x)\implies f(z)\leq (z-x)f(x)+f(f(x)).$

Then $x\to -\infty,$ yields a contradiction. $\blacksquare$

From $(*)$ we get $f(x)=0,\forall x<0.$ $P(0,f(0))\implies f(0)\geq 0,$ thus we get $f(0)=0,$ as desired. $\square$

                           ~ZETA_in_olympiad

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 ==See Also==
-*[[IMO Problems and Solutions]]
+{{IMO box|year=2011|num-b=2|num-a=4}}
 [[Category:Olympiad Algebra Problems]]
 [[Category:Functional Equation Problems]]

2011 IMO (Problems) • Resources
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2011 IMO Problems/Problem 3: Difference between revisions

Latest revision as of 00:20, 19 November 2023

Solution

See Also