Difference between revisions of "2001 IMO Shortlist Problems/A3"
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Hence it is only required to prove <math>a_1^2+a_2^2+................+a_k^2<1</math> where <math>a_k=\dfrac{x_k}{1+x_1^2+... x_k^2}</math> | Hence it is only required to prove <math>a_1^2+a_2^2+................+a_k^2<1</math> where <math>a_k=\dfrac{x_k}{1+x_1^2+... x_k^2}</math> | ||
− | for <math>k \geq 2</math>, <math>a_k^2=(\dfrac{x_k}{1+x_1^2+... x_k^2})^2 | + | for <math>k \geq 2</math>, <math>a_k^2=(\dfrac{x_k}{1+x_1^2+... x_k^2})^2 < \dfrac{x_k^2}{(1+x_1^2+... x_{k-1}^2)(1+x_1^2+... x_k^2)} = \dfrac{1}{1+x_1^2+... x_{k-1}^2}-\dfrac{1}{1+x_1^2+... x_k^2}</math> |
− | For k=1 <math>a_1^2 | + | For k=1, <math>a_1^2 = 1-\dfrac{1}{1+x_1^2}</math> |
Summing these inequalities, the right-hand side yields | Summing these inequalities, the right-hand side yields |
Latest revision as of 00:54, 24 October 2023
Contents
Problem
Let be arbitrary real numbers. Prove the inequality
Solution 1
We prove the following general inequality, for arbitrary positive real :
with equality only when
.
We proceed by induction on . For
, we have trivial equality. Now, suppose our inequality holds for
. Then by inductive hypothesis,
If we let
, then we have
with equality only if
.
By the Cauchy-Schwarz Inequality,
with equality only when
. Since
, our equality cases never coincide, so we have the desired strict inequality for
. Thus our inequality is true by induction. The problem statement therefore follows from setting
.
Solution 2
By the Cauchy-Schwarz Inequality
For all real numbers.
Hence it is only required to prove
where
for ,
For k=1,
Summing these inequalities, the right-hand side yields
Hence Proved by Maths1234RC
P.S. This is my first solution on AOPS.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.