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ohh lol..sorry..it should be >= sign,,wait i'll edit the ques..

done now..

Lemme dust off my real analysis textbook... ahaha...

n! = 1*2*3*...*(n-2)*(n-1)*n
Now,
(1+n/2) = (2+n-1)/2 = (3+n-2)/2

I hope I made it clear

sorry didnt get it :|

where didnt u understand

n! = 1*2*3*...*(n-2)*(n-1)*n

right? @shubhamsrg

so there will be around half number of (1+n)/2 terms.....
+
conclusion part..

yes..following..

Now, take 1 and n and average them

okay,,1+n /2..

So, u will get (n+1)/2 right?

n/2 ..

u will again get (n+1)/2 right?

maybe you meant 2 and n-1 ?

yes,then avg n+1 /2

oh...... 2 and (n-1)

got it?

okay,,i get that part..

buzz ?

guys I think I was totally wrong

But got the method to solve this

am all years..

ears **

@mukushla didnt get you..
@sauravshakya your method ?

go on saurav

take n=2 so n!=2 and (3/2)^2=2.25 so 2<2.25 hence proved

well thats ofcorse not a general soln..

u got n! / (n/2) ! > ((n+1)/2 )^(n/2)
how can we match this with original inequality?

you're the expert,,you telll!! :P

lol...idk if it will help or not.

what were you saying about induction?

hmm

didnt get the second last step..?

well \[1 \le \frac{1}{2} (\frac{n+2}{n+1})^{n+1}\] but u need to prove this inequality also

i see..

u most prove \(\text{something} \le 0\) right?

yep..

i did not read all of the above, but is this supposed to be a proof by induction?

well you can use any legitimate method sir..

no i messed that up, have to try again

people i got the solution!!

it was accidently,,while solving another ques,,and i recalled AM >= GM,,here's my solution

Neat solution

thanks! :)

very nice man...\[\frac{1+2+...+n}{n}\ge\sqrt[n]{n!}\]