Prove Inequality For $x,y,z>0$ When $xyz=1$

[anemone]

For $x,\,y,\,z>0$ and $xyz=1$, prove $\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\le\sqrt{2}(x+y+z)$.

 

[Albert1]

For $x,\,y,\,z>0$ and $xyz=1$, prove $\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\le\sqrt{2}(x+y+z)$.

my solution:

Spoiler

let :$A=\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}---(1)$
and B=$\sqrt{2}(x+y+z)---(2)$
using :Cauchy Schwarg inequalty or $AP\geq GP$ inequality (both will get the same result)
we have :$A^2\leq 3x^2+3y^2+3z^2+9---(3)$
$B^2\leq 6x^2+6y^2+6z^2=3x^2+3y^2+3z^2+3x^2+3y^2+3z^2---(4)$
but $3x^2+3y^2+3z^2\geq 3\sqrt[3]{27x^2y^2z^2}=9 $(using $AP\geq GP )$ again
$\therefore A^2\leq B^2$
this imples $A\leq B$

 

[anemone]

my solution:

Spoiler

let :$A=\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}---(1)$
and B=$\sqrt{2}(x+y+z)---(2)$
using :Cauchy Schwarg inequalty or $AP\geq GP$ inequality (both will get the same result)
we have :$A^2\leq 3x^2+3y^2+3z^2+9---(3)$
$B^2\leq 6x^2+6y^2+6z^2=3x^2+3y^2+3z^2+3x^2+3y^2+3z^2---(4)$
but $3x^2+3y^2+3z^2\geq 3\sqrt[3]{27x^2y^2z^2}=9 $(using $AP\geq GP )$ again
$\therefore A^2\leq B^2$
this imples $A\leq B$

Hi Albert, thanks for participating..

Spoiler

But I don't think you've completed the proof, Albert...

You stated $3x^2+3y^2+3z^2\geq 3\sqrt[3]{27x^2y^2z^2}=9 $, which I agreed that it holds true for $x,\,y,\,z>0$ and $xyz=1$, but I can't see how that is going to help in proving $A\leq B$...sorry Albert...

 

[Albert1]

Hi Albert, thanks for participating..

Spoiler

But I don't think you've completed the proof, Albert...

You stated $3x^2+3y^2+3z^2\geq 3\sqrt[3]{27x^2y^2z^2}=9 $, which I agreed that it holds true for $x,\,y,\,z>0$ and $xyz=1$, but I can't see how that is going to help in proving $A\leq B$...sorry Albert...

Spoiler

compare (3) and (4)
for each pair of x,y,z >0 ,and xyz=1
$A^2 \leq B^2$ always holds
when x=y=z=1 then
$A^2=B^2$ or $A=B$

 

[kaliprasad]

Spoiler

compare (3) and (4)
for each pair of x,y,z >0 ,and xyz=1
$A^2 \leq B^2$ always holds
when x=y=z=1 then
$A^2=B^2$ or $A=B$

I am also not convinced

Spoiler

x < A, y < B even A < B does not mean x < y

for example 3.5 < 4 and 3.4 < 6 does not mean 3.5 < 3.4
x < y need to be proved
in your case $A^2 < B^2$

 

[Albert1]

Spoiler

let us consider the extreme condition :
$\sqrt {x^2+1}-\sqrt 2 x>0---(1)$
$\sqrt {y^2+1}-\sqrt 2 y>0---(2)$
the maximum value of(1)+(2) approaches 2
by using the given condition $xyz=1$
(that is $x ,y$ approach $0^+$ ,and the value of $z$ very big)
if we can find $z$ and $\sqrt 2 z-\sqrt {z^2+1}>2$
then the statement is true
(this can be done when $z$ appoaches $5.1^-$

 

[anemone]

For $x,\,y,\,z>0$ and $xyz=1$, prove $\sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\le\sqrt{2}(x+y+z)$.

Solution of other:

Spoiler

Let $f(x)=\sqrt{2}x-\sqrt{x^2+1}-\dfrac{\sqrt{2}\ln x}{2}$, note that (from graphical method) for $0<x<1$, we have the function $g(x)=\sqrt{2}x-\sqrt{x^2+1}$ lies above the function $h(x)=\dfrac{\sqrt{2}\ln x}{2}$ so $\sqrt{2}x-\sqrt{x^2+1}\ge\dfrac{\sqrt{2}\ln x}{2}$ holds for $0<x\le 1$, equality occurs at $x=1$.

Next, if we could prove $\sqrt{2}x-\sqrt{x^2+1}>\dfrac{\sqrt{2}\ln x}{2}$ for $x>1$, then the result will follow.

By using differentiation method, we find

$f'(x)=\sqrt{2}-\dfrac{x}{\sqrt{x^2+1}}-\dfrac{\sqrt{2}}{2x}=\dfrac{(2x-1)\sqrt{x^2+1}-\sqrt{2}x^2}{x\sqrt{2(x^2+1)}}$,

$f''(x)=\dfrac{1}{\sqrt{2}x^2}-\dfrac{1}{(x^2+1)\sqrt{x^2+1}}$ and $f'(x)=0$ iff $P(x)=(2x-1)\sqrt{x^2+1}-\sqrt{2}x^2=0$.

It's not hard to see $P(1)=0$, and $f''(1)>0$ so $(1,\,0)$ is a minimum point.

Note that $(2x-1)\sqrt{x^2+1}-\sqrt{2}x^2\div (x-1)$ gives $(x-1)(2x^3-2x^2+3x-1)=0$ and since $Q(x)=2x^3-2x^2+3x-1$ is a strictly increasing function beyond $x=1$ we can therefore conclude that $f(x)=\sqrt{2}x-\sqrt{x^2+1}-\dfrac{\sqrt{2}\ln x}{2}\ge 0$ for $x>0$.

Now, for $x,\,y,\,z>0$ and $xyz=1$, we can make the three inequalities as shown below and adding them up gives the desired result.

$\sqrt{2}x-\sqrt{x^2+1}\ge \dfrac{\sqrt{2}\ln x}{2}$

$\sqrt{2}y-\sqrt{y^2+1}\ge \dfrac{\sqrt{2}\ln y}{2}$

$\sqrt{2}z-\sqrt{z^2+1}\ge \dfrac{\sqrt{2}\ln z}{2}$

$\therefore \sqrt{x^2+1}+\sqrt{y^2+1}+\sqrt{z^2+1}\le\sqrt{2}(x+y+z)$.