Quirky Inequality of Functions: f((x^2+ax-2a)/8)>= 4f(1/x)^2-19f(1/x)+25

[Surreal]

The function f() is such that for every x€R and x<0 and for every a>2 the following inequality is fulfilled:
f((x^2+ax-2a)/8)>= 4f(1/x)^2-19f(1/x)+25.
Do numbers m and n and m=!n so f(m)=f(n).
Thanks in advance!
P.S: I tried integrating but it led to nothing of interest. Next I tried prooving by contradiction but other than the fact that the function is of power<=1 I seem to not get anything useful for the solution of this problem.

 

[mmm4444bot]

The function f() …

Is that supposed to be f(x) ?

… Do numbers m and n and m=!n so f(m)=f(n)

Is this a question or an instruction?

We write factorials as: \(\displaystyle n!\)

Is that the meaning of !n ?

 

[Surreal]

Is that supposed to be f(x) ?


Is this a question or an instruction?

We write factorials as: \(\displaystyle n!\)

Is that the meaning of !n ?

I meant to say for every function.

I am new to the formatting. I wanted to express that m isn't equal to n.

 

[Surreal]

I meant to say for every function.

I am new to the formatting. I wanted to express that m isn't equal to n.

Do M and n exist sich as......

 

[Surreal]

A Quirky Inequality of Functions

The*function*f() is such that for every x€R and x<0 and for every a>2 the following inequality is fulfilled:
f((x^2+ax-2a)/8)>= 4f(1/x)^2-19f(1/x)+25.
Do numbers m and n exist such ad m=/=n so f(m)=f(n).
Thanks in advance!
P.S: I tried integrating but it led to nothing of interest. Next I tried prooving by contradiction but other than the fact that the function mustn't be of power€(-1;1). I seem to not get anything useful for the solution of this problem.