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Is x² ≥ α(α-1)?

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,753
Given \(\displaystyle \alpha\) is a non-negative real number and for every real number \(\displaystyle x\), we have \(\displaystyle (x+1)^2\ge \alpha(\alpha+1)\).

Is \(\displaystyle x^2\ge \alpha(\alpha-1)\)?
 

mathworker

Active member
May 31, 2013
118
it might be little fussy,
it is given that for every real number
\(\displaystyle (x+1)^2>=a(a+1)\)
so it can also be written as
\(\displaystyle x^2>=.....\)
as \(\displaystyle a\) is poitive
if \(\displaystyle x^2>=a^2+a\)
then,\(\displaystyle x^2>a^2-a\) but i am not able to prove \(\displaystyle '=' \) part
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
Given \(\displaystyle \alpha\) is a non-negative real number and for every real number \(\displaystyle x\), we have \(\displaystyle (x+1)^2\ge \alpha(\alpha+1)\).

Is \(\displaystyle x^2\ge \alpha(\alpha-1)\)?
we need to check for
x has to be > -1 other wise

x^2 > (x+1)^2 and this condition shall always hold

let us check x> 0 (between 0 and -1 this need to be analysed )O

further if α < x we have the result

so we need to analyse a > x
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
Given \(\displaystyle \alpha\) is a non-negative real number and for every real number \(\displaystyle x\), we have \(\displaystyle (x+1)^2\ge \alpha(\alpha+1)\).

Is \(\displaystyle x^2\ge \alpha(\alpha-1)\)?
Hey anemone! ;)

Here's my attempt.

Since it's true for any $x$, it's also true for $x=-1$.
Therefore \(\displaystyle 0 \ge \alpha(\alpha+1)\).

Since $\alpha \ge 0$, it follows that $\alpha = 0$.

So the requested inequality \(\displaystyle x^2 \ge \alpha(\alpha-1)\) simplifies to \(\displaystyle x^2 \ge 0\), which is indeed true for any real number x. $\qquad \blacksquare$
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
Hey anemone! ;)

Here's my attempt.

Since it's true for any $x$, it's also true for $x=-1$.
Therefore \(\displaystyle 0 \ge \alpha(\alpha+1)\).

Since $\alpha \ge 0$, it follows that $\alpha = 0$.

So the requested inequality \(\displaystyle x^2 \ge \alpha(\alpha-1)\) simplifies to \(\displaystyle x^2 \ge 0\), which is indeed true for any real number x. $\qquad \blacksquare$
I would not take a for alpha to be fixed then the ans becomes trivial (The question might have implied this)
then (x+1)^2 > a(a+1)

and then as a(a+1) >= a(a-1) ( a being positive)

(x+1)^2 >= a(a-1) and as it is true for any x so putitng x -1 as x we get the x.

Now with a which is not fixed

and for ( x+ 1) ^2 > a(a+1) I provide the solution
in next post
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
Given \(\displaystyle \alpha\) is a non-negative real number and for every real number \(\displaystyle x\), we have \(\displaystyle (x+1)^2\ge \alpha(\alpha+1)\).

Is \(\displaystyle x^2\ge \alpha(\alpha-1)\)?
I have dropped the word every and provide the solution for postiive x ( I have already shown for -ve x to be true

we have (x+1)^2≥α(α+1)... (1)

Is x^2≥α(α−1)
we can chose x+1 to be t and have

t^2≥α(α+1)
now as t is positive so t > α

let t = α+h ( h >0)

so t^2 - α(α+1)
= (α+h)^2 - α(α+1)
= 2αh + h^2 - α >= 0 given (1)

we need to show that

(t-1)^2≥α(α-1)

(t-1)^2-α(α-1)
= (α+h-1)^2 - α(α-1)
= (α^2+h^2+1+ 2αh - 2α +2h) - α(α-1)
= h^2+1+ 2αh - α +2h
= ( h^2 + 2αh - α) + ( 1+ 2h)
> 0 as ( h^2 + 2αh - α) from (1)

hence proved that (t-1)^2≥α(α-1