Showing 1/2 3^(2-p)(x+y+z)^(p-1)<=(x^p)/(y+z)+(y^p)/(x+z)+(z^p)/(y+x) By Jensen I got taking p_1=p_2=p_3=1/3, (say x=x_1,y=x_2,z=x_3) (sum p_kx_k)^p<=\sum p_kx^p which means; 3^(1-p)(x+y+z)^p<= x^p+y^p+z^p and if I assume wlog x>=y>=z then 1/(y+z)>=1/(x+z)>=1/(x+y) so these 2 sequences are similarly ordered, and by rearrangement inequality the rest follows ? I should use power means, but any other solution is also appreciated.

Deromediqm 2022-07-21 Answered
Showing 1 2 3 2 p ( x + y + z ) p 1 x p y + z + y p x + z + z p y + x with p > 1, and x , y , z positive
By Jensen I got taking p 1 = p 2 = p 3 = 1 3 , (say x = x 1 , y = x 2 , z = x 3 )
( p k x k ) p p k x p which means;
3 1 p ( x + y + z ) p x p + y p + z p
and if I assume wlog x y z then 1 y + z 1 x + z 1 x + y
so these 2 sequences are similarly ordered, and by rearrangement inequality the rest follows ?
I should use power means, but any other solution is also appreciated.
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (2)

Cael Cox
Answered 2022-07-22 Author has 11 answers
Indeed, Jensen works.
Let x + y + z = 3
Hence, we need to prove that c y c f ( x ) 0, where f ( x ) = x p 3 x 1 2
f ( x ) = x p 2 ( ( p 2 ) ( p 1 ) x 2 6 p ( p 2 ) x + 9 p ( p 1 ) ) ( 3 x ) 3
If p = 2 so f ( x ) > 0
If p > 2 so since f ( 0 ) > 0 so since f ( 0 ) > 0 and lim x 3 f ( x ) > 0 and 3 p p 1 > 3, we see that f ( x ) > 0
If 1 < p < 2 so since f ( 0 ) > 0 and lim x 3 f ( x ) > 0, we see that f ( x ) > 0
Thus, your inequality follows from Jensen.
Done
Did you like this example?
Subscribe for all access
Ibrahim Rosales
Answered 2022-07-23 Author has 7 answers
WLOG,
(1) x + y + z = 1
then by power mean inequality;
c y c ( 1 3 ( x + y ) 1 ) 1 ( x + y ) 1 3 ( y + z ) 1 3 ( x + z ) 1 3
and AM-GM
( x + y ) 1 3 ( y + z ) 1 3 ( x + z ) 1 3 1 3 ( x + y ) + 1 3 ( y + z ) + 1 3 ( x + z ) = ( 1 ) 2 3
1 x + y + 1 y + z + 1 x + z 9 2
Rearrangement Inequality implies,
x p y + z + y p x + z + z p y + x ( x p + y p + z p ) ( 1 x + y + 1 y + z + 1 x + z ) 3
and in general one has ( p k x k ) p p k x p thus,
3 p 1 ( x p + y p + z p ) ( x + y + z ) p = ( 1 ) ( x + y + z ) p 1
x p y + z + y p x + z + z p y + x ( x p + y p + z p ) 9 2 3 3 1 p ( x + y + z ) p 1 9 2 3 = 1 2 3 2 p ( x + y + z ) p 1
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

New questions