Show that x+y+z≥√[((x+1)(y+2)(z+2))/(3)]

[anemone]

Let $x,\,y,\,z$ be positive real numbers satisfying $xyz=1$.

Show that $x+y+z\ge \sqrt{\dfrac{(x+2)(y+2)(z+2)}{3}}$.

 

[Albert1]

Let $x,\,y,\,z$ be positive real numbers satisfying $xyz=1$.

Spoiler

Show that $x+y+z\ge \sqrt{\dfrac{(x+2)(y+2)(z+2)}{3}}---(1)$.

Spoiler

let :$x,y,z $ be the roots of :$ t^3-at^2+bt-1=0 (here \ \, x,y,z>0)$
we have :$a=x+y+z\geq 3---(2)$
and $b=xy+yz+zx\geq 3---(3)$
$xyz=1---(4)$
also $a^2 - 2b \geq 3---(5)$
if (1) is true we only have to prove :$3a^2\geq 9+4a+2b----(*)$
from (2)(3)(5) it is easy to see (*) is true ,
and the proof is done

 

[kaliprasad]

Spoiler

let :$x,y,z $ be the roots of :$ t^3-at^2+bt-1=0 (here \ \, x,y,z>0)$
we have :$a=x+y+z\geq 3---(2)$
and $b=xy+yz+zx\geq 3---(3)$
$xyz=1---(4)$
also $a^2 - 2b \geq 3---(5)$
if (1) is true we only have to prove :$3a^2\geq 9+4a+2b----(*)$
from (2)(3)(5) it is easy to see (*) is true ,
and the proof is done

uanble to follow

Spoiler

$a^2 - 2b \geq 3---(5)$

 

[Albert1]

uanble to follow

Spoiler

$a^2 - 2b \geq 3---(5)$

$a^2-2b=(x+y+z)^2-2(xy+yz+zx)$
$=x^2+y^2+z^2\geq 3\sqrt[3]{x^2y^2z^2}=3$
using:$AP\geq GP$

 

[anemone]

Spoiler

let :$x,y,z $ be the roots of :$ t^3-at^2+bt-1=0 (here \ \, x,y,z>0)$
we have :$a=x+y+z\geq 3---(2)$
and $b=xy+yz+zx\geq 3---(3)$
$xyz=1---(4)$
also $a^2 - 2b \geq 3---(5)$
if (1) is true we only have to prove :$3a^2\geq 9+4a+2b----(*)$
from (2)(3)(5) it is easy to see (*) is true ,
and the proof is done

Thanks Albert for participating and the solution that I want to show you and MHB is more or less using the same method as you:

Spoiler

First, we apply AM-GM to $a^2$ and 1 and we get: $a^2+1\ge 2\sqrt{a^2}\ge 2a$

We repeat this for $b$ and $c$ and add the resulting inequalities yields

$a^2+b^2+c^2+3\ge 2(a+b+c)$---(1)

Next, apply AM-GM to $ab,\,bc,\,ac$ gives us $ab+bc+ac\ge 3\sqrt[3]{(abc)^2}\ge 3$---(2) since $abc=1$.

Again, by AM-GM applying on $a^2,\,b^2,\,c^2$ we have $a^2+b^2+c^2\ge 3\sqrt[3]{(abc)^2}\ge 3$---(3) since $abc=1$.

Now, we add $2\times (1)+4\times (2)+(3)$ and that gives

$3(a^2+b^2+c^2)+4(ab+bc+ac)\ge 4(a+b+c)+9$

Add $2(ab+bc+ac)$ to both sides of the inequality above gives

$3(a^2+b^2+c^2)+6(ab+bc+ac)\ge 2(ab+bc+ac)+4(a+b+c)+9$

$3((a^2+b^2+c^2)+2(ab+bc+ac))\ge (a+2)(b+2)(c+2)$

$3(a+b+c)^2\ge (a+2)(b+2)(c+2)$

$a+b+c\ge \sqrt{\dfrac{(a+2)(b+2)(c+2)}{3}}$ (Q.E.D.)