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

  • Thread starter Thread starter anemone
  • Start date Start date
AI Thread Summary
The discussion centers on the inequality $$x^2 \ge \alpha(\alpha-1)$$ under the condition that $$\alpha$$ is a non-negative real number and for all real $$x$$, $$(x+1)^2 \ge \alpha(\alpha+1)$$ holds true. It is established that if $$\alpha = 0$$, then the inequality simplifies to $$x^2 \ge 0$$, which is valid for any real number $$x$$. Further analysis shows that for positive values of $$x$$, the inequality holds as well, with transformations leading to a proof that confirms the original statement. The conclusion is reached that the inequality is indeed valid under the given conditions.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Given $$\alpha$$ is a non-negative real number and for every real number $$x$$, we have $$(x+1)^2\ge \alpha(\alpha+1)$$.

Is $$x^2\ge \alpha(\alpha-1)$$?
 
Mathematics news on Phys.org
it might be little fussy,
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
 
anemone said:
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
 
Last edited:
anemone said:
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. $\qquad \blacksquare$
 
I like Serena said:
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
 
anemone said:
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
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Back
Top