Albert1
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$a,b,c,m,n,p\in R$
gving :$ap-2bn+cm=0---(1)$
and : $ b^2-ac<0---(2)$
prove :$n^2\geq mp$
gving :$ap-2bn+cm=0---(1)$
and : $ b^2-ac<0---(2)$
prove :$n^2\geq mp$
note that:anemone said:My solution:
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 $n^2\ge mp$.
Albert said: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 :
$a,b,c,m,n,p\in R$