- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,753

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

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,753

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

- May 31, 2013

- 118

\(\displaystyle (x+1)^2>=a(a+1)\)

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

- Mar 31, 2013

- 1,322

we need to check for

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

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:

- Admin
- #4

- Mar 5, 2012

- 8,851

Hey anemone!

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

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. $\qquad \blacksquare$

- Mar 31, 2013

- 1,322

I would not take a for alpha to be fixed then the ans becomes trivial (The question might have implied this)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. $\qquad \blacksquare$

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

- Mar 31, 2013

- 1,322

I have dropped the word every and provide the solution for postiive x ( I have already shown for -ve x to be true

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

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