Harazi ineq

manlio

For $a,b,c$ positive reals such that $ab+bc+ca =3$ prove that

$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{1}{\sqrt{1+c^2}}$

**Posted:** Mon Jan 31, 2005 10:53 am

zhaobin

we can hvae $LHS\geq1,RHS\leq1$ right?

**Posted:** Mon Jan 31, 2005 11:00 am

manlio

Myth said to me that

$LHS \geq 1$

if $ab \leq 3$ and he never makes mistakes Wink

, so it is true that $LHS \geq 1$

but I cannot prove it

**Posted:** Mon Jan 31, 2005 11:10 am

manlio

Perhaps it is better to write Smile

$\sqrt{\frac{1}{1+a^2}}+\sqrt{\frac{1}{1+b^2}} \geq 1$

for $ab \leq 3$

**Posted:** Mon Jan 31, 2005 11:20 am

zhaobin

it is my proof here. let $\frac{1}{\sqrt{1+a^{2}}}=x,\frac{1}{\sqrt{1+b^{2}}}=y$
we can get $(\frac{1}{x^{2}}-1)(\frac{1}{y^{2}}-1)\leq9$ ,then if $x+y<1$
we can have the contradiction

**Posted:** Mon Jan 31, 2005 11:37 am

manlio

You have

$(1-x^2)(1-y^2) \leq 9x^2y^2$ with $x,y \leq 1$

or

$8x^2y^2 +x^2+y^2 \geq 1$

If $x +y <1$

then

$x^2+y^2 +2xy <1$

so

$8x^2y^2 +x^2+y^2 > x^2+y^2 +2xy$

or

$4xy >1$

Why is this a contradiction?

**Posted:** Mon Jan 31, 2005 12:11 pm

zhaobin

$(\frac{1}{x^{2}}-1)(\frac{1}{y^{2}}-1)\leq9$ ,then if $x+y<1$
we have $(\frac{1}{x^{2}}-1)(\frac{1}{y^{2}}-1)= \frac{1-x^{2}}{x^{2}}\frac{1-y^{2}}{y^{2}}>\frac{(x+x+y)y}{x^{2}}\frac{(y+y+x)x}{y...$
use $AM >= GM$

we can have the cantradiction.sorry for my unclear post last time. Smile

**Posted:** Mon Jan 31, 2005 12:47 pm

manlio

Thank you very much , zhaobin.

It is correct and very nice. Smile

**Posted:** Mon Jan 31, 2005 1:22 pm

zhaobin

**Posted:** Mon Jan 31, 2005 2:17 pm

nttu

I have an easier way :
The ineq is equivalent to
$\sqrt{a^2+1} + \sqrt{b^2+1} \geq \sqrt{a^2+1}.\sqrt{b^2+1}$
We have :
$\sqrt{a^2+1} + \sqrt{b^2+1} \geq \sqrt{(a+b)^2 +4}$
We will prove
$\sqrt{(a+b)^2 +4} \geq \sqrt{a^2+1}.\sqrt{b^2+1}$ (1)
Fortunately , $(1) <==> (ab-3) (ab+1) \leq 0$

**Posted:** Tue Feb 01, 2005 1:19 am

_________________
Nguyen Tuan Tu