Is x² ≥ α(α-1)?

anemone

MHB POTW Director
Staff member
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

Well-known member
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

Well-known member
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
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. $\qquad \blacksquare$

Well-known member
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

Well-known member
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