IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

Prove that for any real number x

$\ 2^x + 3^x - 4^x + 6^x - 9^x \leq 1$

konichiwa2x

open to all?

sandeepramesh

case 1: x>0 proved :D

case 2: x<0 proved :D

x=0 implies value is 1. :)

Hence proved :D

outline

for x>0 surely the negative terms exceed the positive terms

group it as (9^x - 6^x - 3^x) for x>=1

for x<0 note that 1/2 + 1/3 + 1/6 =1

computer001

take everything 2 rhs
multiply by 2
express as sum of 3 squares so obviously >=0

hsbhatt

hey i just checked that none of u were online. did u all just wake up?

sandeepramesh

no we were hiding to pounce on you when you give this sum :D :D :D

computer001

the squares will be (3^x-2^x)^2 + (2^x-1)^2 + (3^x-1)^2

konichiwa2x

unfair!

u both didnt even wait for him to say if it was open to all!!

hsbhatt

assuming conditions is one of koni's weaknesses

sandeepramesh

what weakness?? :D :D

well he said for your recreation :D and i requested a sum from him so i could try :D :P :P

hsbhatt

If you let $a = 2^x ext {and} b = 3^x$

Then we are asked to prove that a+b+ab leq a^2+b^2+1

This is true as $1+a^2 geq 2a, 1+b^2 geq 2b ext{and} a^2+b^2 geq 2ab$

Ok that is a strict inequality

PS: I could have just said use the inequality

$p^2 + q^2 + r^2 \geq pq+qr+rp$

konichiwa2x

For the case

,

and so on...

In general,

hence inequality simplifies to

i.e

for x > 1 which is obviosully true.

Similar method for x < 1 , with the differnce that

.

konichiwa2x

nice sir... the above is the method i had in mind.

sandeep, last time he said its not open to everyone..so just guess it might be the same this time too.

Anyway, i wont let this weakness let me down again

sandeepramesh

well i requested it :P