Let $a,b,c,d,e,f$ be positive numbers such that $a,b,c,d$ is an arithmetic progression, and $a,e,f,d$ is a geometric progression. Prove that $bc\ge ef$.
Problem
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Tags: inequalities
Yeetopedia
07.04.2021 17:57
Let $D$ we the common difference of the arithmetic sequence and $g$ be the commonm difference of the geometric sequence(is that what’s it called sorry I forgot). Let’s write everything in terms of $a$ and $d$.
$b=a+D$.
$c=a+2D$.
Now what we can write in t of $a$ and $g$:
$e=ag$
$f=2ag$
Also, looking at the sequence, we know that $eg,e,f \leq d$, and $fg=d=3ag$ So, $e+f=d$.
Writing $d$ in terms of $D$, we get:
$a+3D=c+D=b+2D$.
So, $c+b=d$.
So, $bc$ can be equal to $ef$.
Multiplying what we had wrote above, we get:
$a^2+2D^2+3Da?2(ag)^2$
We know that $ag=e$, so
$a^2+2D^2+3Da?2e^2$.
However, using the fact that $b+c=d$, we can substitute in $>$ in our equation above.
So, $bc \geq ef$
$\square$
Click to reveal hidden textIdk how right this is, I think it might be a litttle flawed would appreciate where I can fix it
Yeetopedia
08.04.2021 16:06
Yeetopedia wrote:
Let $D$ we the common difference of the arithmetic sequence and $g$ be the commonm difference of the geometric sequence(is that what’s it called sorry I forgot). Let’s write everything in terms of $a$ and $d$.
$b=a+D$.
$c=a+2D$.
Now what we can write in t of $a$ and $g$:
$e=ag$
$f=2ag$
Also, looking at the sequence, we know that $eg,e,f \leq d$, and $fg=d=3ag$ So, $e+f=d$.
Writing $d$ in terms of $D$, we get:
$a+3D=c+D=b+2D$.
So, $c+b=d$.
So, $bc$ can be equal to $ef$.
Multiplying what we had wrote above, we get:
$a^2+2D^2+3Da?2(ag)^2$
We know that $ag=e$, so
$a^2+2D^2+3Da?2e^2$.
However, using the fact that $b+c=d$, we can substitute in $>$ in our equation above.
So, $bc \geq ef$
$\square$
Click to reveal hidden textIdk how right this is, I think it might be a litttle flawed would appreciate where I can fix it
could someone check this plz?
RedFireTruck
08.04.2021 16:10
Yeetopedia wrote:
Let $D$ we the common difference of the arithmetic sequence and $g$ be the commonm difference of the geometric sequence(is that what’s it called sorry I forgot). Let’s write everything in terms of $a$ and $d$.
$b=a+D$.
$c=a+2D$.
Now what we can write in t of $a$ and $g$:
$e=ag$
$f=2ag$
Also, looking at the sequence, we know that $eg,e,f \leq d$, and $fg=d=3ag$ So, $e+f=d$.
Writing $d$ in terms of $D$, we get:
$a+3D=c+D=b+2D$.
So, $c+b=d$.
So, $bc$ can be equal to $ef$.
Multiplying what we had wrote above, we get:
$a^2+2D^2+3Da?2(ag)^2$
We know that $ag=e$, so
$a^2+2D^2+3Da?2e^2$.
However, using the fact that $b+c=d$, we can substitute in $>$ in our equation above.
So, $bc \geq ef$
$\square$
Click to reveal hidden textIdk how right this is, I think it might be a litttle flawed would appreciate where I can fix it
g is the common ratio, and $f=ag^2$ not $2ag$
kootrapali
08.04.2021 16:23
Since $b+c=a+d$ and $b,c$ are closer to each other or equal, $bc\geq ad$. But $ad=ef$, so $bc\geq ef$, as desired.
Taco12
30.11.2021 00:34
From the information given, we find $\frac{d}{f}=\frac{f}{e}=\frac{e}{a}$ and $d-c=c-b=b-a$. Also, we have $ad=ef$, so the problem reduces to showing $bc \geq ad$. From the arithmetic sequence, we find $a+d=b+c$, but we also know $c-b>d-a$, so we can conclude that $bc \geq ad$, as desired.