MHB Is x² ≥ α(α-1) Given Certain Conditions for Non-negative Real Numbers?

[anemone]

Given $$\alpha$$ is a non-negative real number and for every real number $$x$$, we have $$(x+1)^2\ge \alpha(\alpha+1)$$.

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

 

[mathworker]

it might be little fussy,

Spoiler

it is given that for every real number
$$(x+1)^2>=a(a+1)$$
so it can also be written as
$$x^2>=...$$
as $$a$$ is poitive
if $$x^2>=a^2+a$$
then,$$x^2>a^2-a$$ but i am not able to prove $$'=' $$ part

 

[kaliprasad]

Given $$\alpha$$ is a non-negative real number and for every real number $$x$$, we have $$(x+1)^2\ge \alpha(\alpha+1)$$.

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

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

 

[I like Serena]

Given $$\alpha$$ is a non-negative real number and for every real number $$x$$, we have $$(x+1)^2\ge \alpha(\alpha+1)$$.

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

Hey anemone! ;)

Here's my attempt.

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

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

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

 

[kaliprasad]

Hey anemone! ;)

Here's my attempt.

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

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

So the requested inequality $$x^2 \ge \alpha(\alpha-1)$$ simplifies to $$x^2 \ge 0$$, 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]

Given $$\alpha$$ is a non-negative real number and for every real number $$x$$, we have $$(x+1)^2\ge \alpha(\alpha+1)$$.

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

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