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#1 h Sep 18, 2010, 12:56 PM • 1 Y

Y by Adventure10

find the minimum value of :
$F= 2 \tan^{2}\frac{A}{2} + 3 \tan^{2}\frac{B}{2} +4\tan^{2}\frac{C}{2}$ if $A, B, C$ be the three angles of triangle $ABC$

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#2 h Sep 19, 2010, 8:43 AM • 2 Y

Y by Adventure10, Mango247

huyhoang wrote:

find the minimum value of :
$F= 2 \tan^{2}\frac{A}{2} + 3 \tan^{2}\frac{B}{2} +4\tan^{2}\frac{C}{2}$ if $A, B, C$ be the three angles of triangle $ABC$

because ABC is a triangle, so we can let $\tan \frac{A}{2} = \sqrt {\frac{{yz}}{{x\left( {x + y + z} \right)}}} ,x,y,z > 0,......$ then

$F = \frac{{2y^2 z^2 + 3z^2 x^2 + 4x^2 y^2 }}{{xyz\left( {x + y + z} \right)}}$ ;

let $yz = p,zx = q,xy = r,p,q,r > 0$ , then

$F = \frac{{2p^2 + 3q^2 + 4r^2 }}{{pq + qr + rp}}$ ;

am-gm

$pq + qr + rp = \sqrt m p \cdot \frac{q}{{\sqrt m }} + \sqrt n q \cdot \frac{r}{{\sqrt n }} + \sqrt l r \cdot \frac{p}{{\sqrt l }}$ $\le \frac{{\left( {m + \frac{1}{l}} \right)p^2 + \left( {n + \frac{1}{m}} \right)q^2 + \left( {l + \frac{1}{n}} \right)r^2 }}{2}$ ,

so we need

$\left( {m + \frac{1}{l}} \right):\left( {n + \frac{1}{m}} \right):\left( {l + \frac{1}{n}} \right) = 2:3:4$ ,

and the equality holds need

$\sqrt m p = \frac{q}{{\sqrt m }},\sqrt n q = \frac{r}{{\sqrt n }},\sqrt l r = \frac{p}{{\sqrt l }} \Rightarrow mnl = 1$ ,

so we need to solve

$\left( {m + mn} \right):\left( {n + \frac{1}{m}} \right):\left( {\frac{1}{{mn}} + \frac{1}{n}} \right) = 2:3:4$ .

to be continue....

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#3 h Sep 20, 2010, 4:00 AM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

huyhoang wrote:

find the minimum value of :
$F= 2 \tan^{2}\frac{A}{2} + 3 \tan^{2}\frac{B}{2} +4\tan^{2}\frac{C}{2}$ if $A, B, C$ be the three angles of triangle $ABC$

because ABC is a triangle, so we can let $\tan \frac{A}{2} = \sqrt {\frac{{yz}}{{x\left( {x + y + z} \right)}}} ,x,y,z > 0,......$ then

$F = \frac{{2y^2 z^2 + 3z^2 x^2 + 4x^2 y^2 }}{{xyz\left( {x + y + z} \right)}}$ ;

let $yz = p,zx = q,xy = r,p,q,r > 0$ , then

$F = \frac{{2p^2 + 3q^2 + 4r^2 }}{{pq + qr + rp}}$ ;

am-gm

$pq + qr + rp = \sqrt m p \cdot \frac{q}{{\sqrt m }} + \sqrt n q \cdot \frac{r}{{\sqrt n }} + \sqrt l r \cdot \frac{p}{{\sqrt l }}$ $\le \frac{{\left( {m + \frac{1}{l}} \right)p^2 + \left( {n + \frac{1}{m}} \right)q^2 + \left( {l + \frac{1}{n}} \right)r^2 }}{2}$ ,

so we need

$\left( {m + \frac{1}{l}} \right):\left( {n + \frac{1}{m}} \right):\left( {l + \frac{1}{n}} \right) = 2:3:4$ ,

and the equality holds need

$\sqrt m p = \frac{q}{{\sqrt m }},\sqrt n q = \frac{r}{{\sqrt n }},\sqrt l r = \frac{p}{{\sqrt l }} \Rightarrow mnl = 1$ ,

so we need to solve

$\left( {m + mn} \right):\left( {n + \frac{1}{m}} \right):\left( {\frac{1}{{mn}} + \frac{1}{n}} \right) = 2:3:4$ .

to be continue....

my solution only use trigonometry inequality

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#4 h Sep 20, 2010, 12:29 PM • 2 Y

Y by Adventure10, Mango247

can you post your prove here? thanks ...

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#5 h Sep 21, 2010, 1:04 PM • 2 Y

Y by Adventure10, Mango247

it based on this inequality:
for all $A, B, C \in R$ such that $\frac{A+B+C}{\pi}$ is an odd integer.
so that for all $x, y, z \in R$ we have the following inequality:
$2xy\cos C+ 2yz\cos A+ 2zx\cos B \le x^{2}+ y^{2}+ z^{2}$

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#6 h Sep 22, 2010, 5:32 AM • 2 Y

Y by Adventure10, Mango247

huyhoang wrote:

it based on this inequality:
for all $A, B, C \in R$ such that $\frac{A+B+C}{\pi}$ is an odd integer.
so that for all $x, y, z \in R$ we have the following inequality:
$2xy\cos C+ 2yz\cos A+ 2zx\cos B \le x^{2}+ y^{2}+ z^{2}$

for $\tan ^2 \frac{A}{2} = \frac{{1 - \cos A}}{{1 + \cos A}}$ then F becomes

$F = 2 \cdot \frac{{1 - \cos A}}{{1 + \cos A}} + 3 \cdot \frac{{1 - \cos B}}{{1 + \cos B}} + 4 \cdot \frac{{1 - \cos C}}{{1 + \cos C}}$

then how to use that inequality ?

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#7 h Sep 22, 2010, 3:32 PM • 1 Y

Y by Adventure10

it so easy!
you just consider $\tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2}$ as $x, y, z$ in my inequality
and consider $\Delta XYZ$ satisfy the following inequality.
Note that $\Sigma \tan\frac{A}{2}\tan\frac{B}{2} =1$

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#8 h Sep 22, 2010, 4:45 PM • 1 Y

Y by Adventure10

huyhoang wrote:

it so easy!
you just consider $\tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2}$ as $x, y, z$ in my inequality
and consider $\Delta XYZ$ satisfy the following inequality.
Note that $\Sigma \tan\frac{A}{2}\tan\frac{B}{2} =1$

oh thanks, I got your mean, it is the following?:

for $\triangle XYZ$ and $x,y,z\in R$ , we have
$2xy\cos Z + 2yz\cos X + 2zx\cos Y \le x^2 + y^2 + z^2$
then let $x = \sqrt 2 \tan \frac{A}{2},y = \sqrt 3 \tan \frac{B}{2},z = 2\tan \frac{C}{2}$ , get

$2\sqrt 6 \tan \frac{A}{2}\tan \frac{B}{2}\cos Z + 4\sqrt 3 \tan \frac{B}{2}\tan \frac{C}{2}\cos X + 4\sqrt 2 \tan \frac{C}{2}\tan \frac{A}{2}\cos Y$ $\le 2\tan ^2 \frac{A}{2} + 3\tan ^2 \frac{B}{2} + 4\tan ^2 \frac{C}{2} = F$

so we need to find a $\triangle XYZ$ such that
$2\sqrt 6 \cos Z = 4\sqrt 3 \cos X = 4\sqrt 2 \cos Y$ . (1)
but this trigonometry equation is not easy to solve (I think

), still to be continue.
and I worry about one thing, that's the equality holds if and only if $x:y:z=\sin X : \sin Y : \sin Z$ , i.e.
$\sqrt 2 \tan \frac{A}{2}:\sqrt 3 \tan \frac{B}{2}:2\tan \frac{C}{2}=\sin X : \sin Y : \sin Z$ . (2)
when we found the $\triangle XYZ$ satisfy (1), is it exist a $\triangle ABC$ satisfy (2)?

thanks again...

sleep... tomorrow continue...

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#9 h Sep 23, 2010, 2:23 AM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

huyhoang wrote:

it so easy!
you just consider $\tan \frac{A}{2}, \tan \frac{B}{2}, \tan \frac{C}{2}$ as $x, y, z$ in my inequality
and consider $\Delta XYZ$ satisfy the following inequality.
Note that $\Sigma \tan\frac{A}{2}\tan\frac{B}{2} =1$

oh thanks, I got your mean, it is the following?:

for $\triangle XYZ$ and $x,y,z\in R$ , we have
$2xy\cos Z + 2yz\cos X + 2zx\cos Y \le x^2 + y^2 + z^2$
then let $x = \sqrt 2 \tan \frac{A}{2},y = \sqrt 3 \tan \frac{B}{2},z = 2\tan \frac{C}{2}$ , get

$2\sqrt 6 \tan \frac{A}{2}\tan \frac{B}{2}\cos Z + 4\sqrt 3 \tan \frac{B}{2}\tan \frac{C}{2}\cos X + 4\sqrt 2 \tan \frac{C}{2}\tan \frac{A}{2}\cos Y$ $\le 2\tan ^2 \frac{A}{2} + 3\tan ^2 \frac{B}{2} + 4\tan ^2 \frac{C}{2} = F$

so we need to find a $\triangle XYZ$ such that
$2\sqrt 6 \cos Z = 4\sqrt 3 \cos X = 4\sqrt 2 \cos Y$ . (1)
but this trigonometry equation is not easy to solve (I think

), still to be continue.
and I worry about one thing, that's the equality holds if and only if $x:y:z=\sin X : \sin Y : \sin Z$ , i.e.
$\sqrt 2 \tan \frac{A}{2}:\sqrt 3 \tan \frac{B}{2}:2\tan \frac{C}{2}=\sin X : \sin Y : \sin Z$ . (2)
when we found the $\triangle XYZ$ satisfy (1), is it exist a $\triangle ABC$ satisfy (2)?

thanks again...

sleep... tomorrow continue...

yes! you're right

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#10 h Sep 23, 2010, 7:24 AM • 2 Y

Y by Adventure10, Mango247

from the identity $\cos ^2 X + \cos ^2 Y + \cos ^2 Z + 2\cos X\cos Y\cos Z = 1$ .

and let $2\sqrt 6 \cos Z = 4\sqrt 3 \cos X = 4\sqrt 2 \cos Y = t$ . (easy to know $t>0$ )

then we can get that $t^3 + 9t^2 - 96 = 0$ .

this equation have only one positive root $t=t_0 \approx 2.84667$ ...

then we can say $F\ge t_0\sum{\tan\frac{A}{2}\tan\frac{B}{2}}=t_0$ , when dose the equality holds?
i.e. : is the following equation set have roots ?
$\left\{ \begin{array}{l} \sqrt 2 \tan \frac{A}{2}:\sqrt 3 \tan \frac{B}{2}:2\tan \frac{C}{2} = \sin X:\sin Y:\sin Z \\ \sum {\tan \frac{A}{2}\tan \frac{B}{2}} = 1 \\ \end{array} \right.$
I think it's sure...

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