[/quote]
,
prove by Cauchy Inequality?
and using the integer 
to arrive at a contradiction.
be distinct positive integers. Prove that
Further, determine when equality holds.
be a positive integer. Find the smallest positive integer 
such that for any set 
of 
such that for all 
and 
then 
Prove that
Let 
Prove that
Let 
Prove that
Let 
Prove that
Prove that
,
Prove that
Let 
Prove that
Let 
Prove that
Let 
Prove that
be the 
a
and 
be foots of perpendicular lines through 
on 
,
and altitudes 
of the triangle 
lies on 
)
such that 
Y by
Suppose a function 
satisfies 
for all real 
. Then, we have 
We also have ![\[f(\tfrac{x+y}{2}) + f(\tfrac{x-y}{2}) \ge 2f(x).\]](//latex.artofproblemsolving.com/5/3/d/53d5e6a96b03a5c4a05a9da1945cbaa81829271e.png)
Adding the first two inequalities and using the third, we have ![\[2f(x) + f(y)+f(-y) \ge 2(f(\tfrac{x+y}{2}) + f(\tfrac{x-y}{2})) \ge 4f(x),\]](//latex.artofproblemsolving.com/a/9/c/a9c9255deeb88e31603a55d6cf32fbf864ca0758.png)
which implies 
. But this means is bounded above which is absurd -- take the identity function. I feel I am making some blithely stupid mistake, but I can't see what. Any help is appreciated 
satisfies 
for all real 
. Then, we have 
We also have ![\[f(\tfrac{x+y}{2}) + f(\tfrac{x-y}{2}) \ge 2f(x).\]](http://latex.artofproblemsolving.com/5/3/d/53d5e6a96b03a5c4a05a9da1945cbaa81829271e.png)
Adding the first two inequalities and using the third, we have ![\[2f(x) + f(y)+f(-y) \ge 2(f(\tfrac{x+y}{2}) + f(\tfrac{x-y}{2})) \ge 4f(x),\]](http://latex.artofproblemsolving.com/a/9/c/a9c9255deeb88e31603a55d6cf32fbf864ca0758.png)
which implies 
. But this means is bounded above which is absurd -- take the identity function. I feel I am making some blithely stupid mistake, but I can't see what. Any help is appreciated This post has been edited 3 times. Last edited by Sedro, Jun 17, 2024, 3:57 PM
on the RHS.
Thanks.
