MHB Proof of $a^{2a}\times b^{2b}\times c^{2c}>a^{b+c}\times b^{c+a}\times c^{a+b}$

[Albert1]

given :
$a>b>c>0$
prove :
$a^{2a}\times b^{2b}\times c^{2c}>a^{b+c}\times b^{c+a}\times c^{a+b}$

 

[Albert1]

given :
$a>b>c>0$
prove :
$a^{2a}\times b^{2b}\times c^{2c}>a^{b+c}\times b^{c+a}\times c^{a+b}$

hint:

Spoiler

let:
$A=a^{2a}\times b^{2b}\times c^{2c}$
$B=a^{b+c}\times b^{c+a}\times c^{a+b}$
prove:$\dfrac {A}{B}>1$

 

[Albert1]

hint:

Spoiler

let:
$A=a^{2a}\times b^{2b}\times c^{2c}$
$B=a^{b+c}\times b^{c+a}\times c^{a+b}$
prove:$\dfrac {A}{B}>1$

Spoiler

for $a>b>c>0$ we have :
$\dfrac {A}{B}=a^{(a-b)}a^{(a-c)}b^{(b-a)}b^{(b-c)}c^{(c-a)}c^{(c-b)}$
$=(\dfrac{a}{b})^{a-b}(\dfrac{b}{c})^{b-c}(\dfrac{a}{c})^{a-c}>1^0\times1^0\times 1^0=1$