my inequality4

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my inequality4

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#1 h Jun 8, 2011, 2:35 AM • 2 Y

Y by Adventure10, Mango247

for a,b,c>0 such that ab+bc+ca=1 prove the following inequality
$\frac{1}{{\sqrt {{x^3} + x} }} + \frac{1}{{\sqrt {{y^3} + y} }} + \frac{1}{{\sqrt {{z^3} + z} }} \ge \frac{9}{{2\sqrt {x + y + z} }}$

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Pantum

#2 h Jun 8, 2011, 3:46 AM • 1 Y

Y by Adventure10

Should be $xy+yz+zx=1,x,y,z>0$ !

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#3 h Jun 8, 2011, 6:35 AM • 1 Y

Y by Adventure10

Pantum wrote:

Should be $xy+yz+zx=1,x,y,z>0$ !

yes so sincerely apologize for this mistake

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#4 h Jun 8, 2011, 8:05 AM • 2 Y

Y by Adventure10, Mango247

Wrong solution
Click to reveal hidden text

However i shall try to do this later on, so to be continued!

Let m=x+y
p=x+z
q=y+z
then we have
$\sum_{cyc} \frac{1}{\sqrt{(m+p-q)(m)(p)}} \geq \frac{9}{{2\sqrt{m+p+q}}}$

Next without loosing generality i can assume that m,p,q are sides of a triangle where
$2(mp + pq + mq)-(m^2 + p ^2 + q ^2)=1$
meaning
$(m+p+q)^2=1+2(m^2 + p ^2 + q ^2)$
To be continued after school

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#5 h Jun 8, 2011, 1:51 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Moderator edit: do not quote the whole post above
thank you very much and this is my proof
$\frac{1}{{\sqrt {{x^3} + x} }} + \frac{1}{{\sqrt {{y^3} + y} }} + \frac{1}{{\sqrt {{z^3} + z} }} = \frac{1}{{\sqrt {x(x + y)(x + z)} }} + \frac{1}{{\sqrt {y(y + z)(y + x)} }} + \frac{1}{{\sqrt {z(z + x)(z + y)} }}$
Thus we must prove
$\frac{1}{{\sqrt {x(x + y)(x + z)} }} + \frac{1}{{\sqrt {y(y + z)(y + x)} }} + \frac{1}{{\sqrt {z(z + x)(z + y)} }} \ge \frac{9}{{2\sqrt {x + y + z} }}$
$\Leftrightarrow \sum {\sqrt {\frac{{xy}}{{(x + z)(y + z)}}} } \ge \frac{{9\sqrt {xyz} }}{{2\sqrt {x + y + z} }}$
apply the familiar inequality
$\sum {\sqrt {\frac{{xy}}{{(x + z)(y + z)}}} } \ge 1 + \frac{{4xyz}}{{(x + y)(y + z)(z + x)}}$
therefore, we must prove
$1 + \frac{{4xyz}}{{(x + y)(y + z)(z + x)}} \ge \frac{{9\sqrt {xyz} }}{{2\sqrt {x + y + z} }}$
$\Leftrightarrow 1 + \frac{4}{{\frac{{x + y + z}}{{xyz}} - 1}} \ge \frac{{9\sqrt {xyz} }}{{2\sqrt {x + y + z} }}$
set $\sqrt {\frac{{x + y + z}}{{xyz}}} = t(t \ge 3)$
$\Leftrightarrow (t - 3)(2{t^2} - 3t - 3) \ge 0$
inequality proved

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#6 h Jun 8, 2011, 5:16 PM • 2 Y

Y by Adventure10, Mango247

$\sum{\sqrt{\frac{{xy}}{{(x+z)(y+z)}}}}\ge 1+\frac{{4xyz}}{{(x+y)(y+z)(z+x)}}$

I would like to know how did you come and what is the proof to this part(I'm new to inequalities and a bit more advanced maths so I'm still in the process of learning). Thank you in advance

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231 posts

#7 h Jun 9, 2011, 3:26 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

It can be written as
$\sum_{cyc}\frac{(x+y)\sqrt{xy(z+x)(z+y)}}{(x+y)(y+z)(z+x)}\ge 1+\frac{4xyz}{(x+y)(y+z)(z+x)}$
use AM-GM
$(x+y)\sqrt{xy(z+x)(z+y)}\ge (x+y)\sqrt{xy}(z+\sqrt{xy})=xy(x+y)+z(x+y)\sqrt{xy}\ge xy(x+y)+2xyz$
then
$\sum_{cyc}\frac{(x+y)\sqrt{xy(z+x)(z+y)}}{(x+y)(y+z)(z+x)}\ge\frac{xy(x+y)+yz(y+z)+zx(z+x)+6xyz}{(x+y)(y+z)(z+x)}=1+\frac{4xyz}{(x+y)(y+z)(z+x)}$

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#8 h Jun 9, 2011, 10:23 AM • 2 Y

Y by Adventure10, Mango247

Thank you guiguzi

By the way do you know the exact name of the inequality?

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gaussintraining

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gaussintraining

#9 h Jun 12, 2011, 4:23 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Per the conditions of the problem statement, we may substitute $x=\tan\frac{A}{2}$ , $y=\tan\frac{B}{2}$ and $z=\tan\frac{C}{2}$ for $\triangle ABC$ . Therefore, the inequality is equivalent to $\sum\frac{\cos\frac{A}{2}}{\sqrt{\tan\frac{A}{2}}}\ge\frac92\left(\frac{1}{\sqrt{\sum\tan\frac{A}{2}}}\right)$ .

BY AM-GM, $\sum\frac{\cos\frac{A}{2}}{\sqrt{\tan\frac{A}{2}}}\ge3\sqrt[3]{\frac{\prod\cos\frac{A}{2}}{\prod\sqrt{\tan\frac{A}{2}}}}$ .

Therefore using the identities: $\begin{cases}\prod\cos\frac{A}{2}=\frac{s}{4R}\\ \prod\tan\frac{A}{2}=\frac{r}{s}\\ \sum\tan\frac{A}{2}=\frac{r+4R}{s}\end{cases}$ , we now have to prove

$\sqrt[3]{\frac{(s)^{\frac32}}{4R\sqrt{r}}}\ge\frac32\left(\frac{\sqrt{s}}{\sqrt{r+4R}}\right)$

$\implies \sqrt[3]{\frac{1}{4R\sqrt{r}}}\ge\frac32\left(\frac{1}{\sqrt{r+4R}}\right)$

$\implies \frac49(r+4R)\ge\sqrt[3]{16R^2r}$

$\implies \frac{64}{729}(r^3+48R^2r+12Rr^2+64R^3)\ge16R^2r$

$\implies 4r^2-537R^2r+48Rr^2+256R^3\ge0$

$\implies (R-2r)(256R^2-25Rr-2r^2)\ge0$

which is clearly true.

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#10 h Jun 13, 2011, 7:29 AM • 2 Y

Y by Adventure10, Mango247

sir_hoang wrote:

Moderator edit: do not quote the whole post above
thank you very much and this is my proof

set $\sqrt {\frac{{x + y + z}}{{xyz}}} = t(t \ge 3)$

inequality proved

??????

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#11 h Jun 13, 2011, 8:21 AM • 2 Y

Y by Adventure10, Mango247

choisiwon wrote:

sir_hoang wrote:

Moderator edit: do not quote the whole post above
thank you very much and this is my proof

set $\sqrt {\frac{{x + y + z}}{{xyz}}} = t(t \ge 3)$

inequality proved

??????

why not, you

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#12 h Jun 13, 2011, 8:29 AM • 1 Y

Y by Adventure10

Lemma has many applications in
for a,b,c $\ge 0$ prove that
$a,\sqrt {\frac{{ab}}{{(a + c)(b + c)}}} + \sqrt {\frac{{bc}}{{(b + a)(c + a)}}} + \sqrt {\frac{{ca}}{{(c + b)(a + b)}}} \ge 1 + \frac{{9abc}}{{2(a + b + c)(ab + bc + ca)}}$
$b,\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2\sqrt {1 + \frac{{abc}}{{(a + b)(b + c)(c + a)}}}$

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#13 h Jun 15, 2011, 5:53 AM • 2 Y

Y by Adventure10, Mango247

sir_hoang wrote:

$b,\sqrt {\frac{a}{{b + c}}} + \sqrt {\frac{b}{{c + a}}} + \sqrt {\frac{c}{{a + b}}} \ge 2\sqrt {1 + \frac{{abc}}{{(a + b)(b + c)(c + a)}}}$

See here http://mathifc.wordpress.com/2010/07/14/inequality-32-2/

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