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

  • Thread starter Thread starter Albert1
  • Start date Start date
  • Tags Tags
    Inequality Proof
AI Thread Summary
The discussion centers on proving the inequality \( a^{2a} \times b^{2b} \times c^{2c} > a^{b+c} \times b^{c+a} \times c^{a+b} \) under the condition that \( a > b > c > 0 \). Participants explore various mathematical approaches and techniques to establish this inequality, emphasizing the significance of the relationship between the variables. Several hints and strategies are proposed, including the application of logarithmic properties and inequalities. The conversation highlights the complexity of the proof and the need for a rigorous mathematical foundation. Ultimately, the goal is to validate the inequality through logical reasoning and mathematical principles.
Albert1
Messages
1,221
Reaction score
0
given :
$a>b>c>0$
prove :
$a^{2a}\times b^{2b}\times c^{2c}>a^{b+c}\times b^{c+a}\times c^{a+b}$
 
Mathematics news on Phys.org
Albert said:
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:
let:
$A=a^{2a}\times b^{2b}\times c^{2c}$
$B=a^{b+c}\times b^{c+a}\times c^{a+b}$
prove:$\dfrac {A}{B}>1$
 
Albert said:
hint:
let:
$A=a^{2a}\times b^{2b}\times c^{2c}$
$B=a^{b+c}\times b^{c+a}\times c^{a+b}$
prove:$\dfrac {A}{B}>1$
for $a>b>c>0$ we have :
$\dfrac {A}{B}=a^{(a-b)}a^{(a-c)}b^{(b-a)}b^{(b-c)}c^{(c-a)}c^{(c-b)}$
$=(\dfrac{a}{b})^{a-b}(\dfrac{b}{c})^{b-c}(\dfrac{a}{c})^{a-c}>1^0\times1^0\times 1^0=1$
 
Last edited:
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top