Proof of $n^2 \geq mp$ from $(1)$ and $(2)$

[Albert1]

$a,b,c,m,n,p\in R$
gving :$ap-2bn+cm=0---(1)$
and : $ b^2-ac<0---(2)$
prove :$n^2\geq mp$

 

[anemone]

My solution:

Spoiler

From $ap-2bn+cm=0$, we get $ap+cm=2bn$, AM-GM inequality tells us $\dfrac{ap+cm}{2}\ge \sqrt{apcm}$ and this is equivalent to $\dfrac{2bn}{2}\ge \sqrt{apcm}$, or simply, $b^2\ge \dfrac{apcm}{n^2}$---(1)

We are also given that $b^2<ac$---(2). Combining these two inequalities (1) and (2) we see that

$ac\ge \dfrac{apcm}{n^2}$

$ac\left(1-\dfrac{pm}{n^2}\right) \ge 0$

Since $ac>0$ which we concluded from $b^2<ac$, we can now say $1-\dfrac{pm}{n^2}\ge 0$ and this leads us to $n^2\ge mp$.

 

[Albert1]

My solution:

Spoiler

From $ap-2bn+cm=0$, we get $ap+cm=2bn$, AM-GM inequality tells us $\dfrac{ap+cm}{2}\ge \sqrt{apcm}$ and this is equivalent to $\dfrac{2bn}{2}\ge \sqrt{apcm}$, or simply, $b^2\ge \dfrac{apcm}{n^2}$---(1)

We are also given that $b^2<ac$---(2). Combining these two inequalities (1) and (2) we see that

$ac\ge \dfrac{apcm}{n^2}$

$ac\left(1-\dfrac{pm}{n^2}\right) \ge 0$

Since $ac>0$ which we concluded from $b^2<ac$, we can now say $1-\dfrac{pm}{n^2}\ge 0$ and this leads us to $n^2\ge mp$.

note that:
AM-GM inequality tells us $\dfrac{ap+cm}{2}\ge \sqrt{apcm}$
we must have :$a,p,c,m>0$
or $a,p,c,m\in R^+$
or at least :
$ap>0 ,\,\, and ,\, cm>0$
but we are given :
$a,b,c,m,n,p\in R$

 

[kaliprasad]

Spoiler

We have 2bn = ap + cm
Square both sides $4b^2n^2 = (ap+cm)^2 = (ap-cm)^2 + 4apcm$
So $4b^2n^2 \ge 4apcm$
as $ac \gt b^2$ as given
so $4acn^2 \gt 4b^2n^2 \ge 4apcm$ as $n^2$ and $ac$ both are $\ge 0$
as $ac > b^2$ so positive hence dividing by 4ac we get the result

 

[anemone]

note that:
AM-GM inequality tells us $\dfrac{ap+cm}{2}\ge \sqrt{apcm}$
we must have :$a,p,c,m>0$
or $a,p,c,m\in R^+$
or at least :
$ap>0 ,\,\, and ,\, cm>0$
but we are given :
$a,b,c,m,n,p\in R$

Ops......

What a glaring error that I made while I used the AM-GM inequality formula...sorry Albert, please disregard my silly solution.

Btw, kaliprasad, your solution is great! (Yes)