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

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    Inequality Proof
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SUMMARY

The inequality \( a^{2a} \times b^{2b} \times c^{2c} > a^{b+c} \times b^{c+a} \times c^{a+b} \) holds true under the condition \( a > b > c > 0 \). This conclusion is derived from applying the AM-GM inequality and properties of exponential functions. The proof involves manipulating the terms and leveraging the relationships between \( a \), \( b \), and \( c \) to establish the desired inequality definitively.

PREREQUISITES
  • Understanding of exponential functions and inequalities
  • Familiarity with the AM-GM (Arithmetic Mean-Geometric Mean) inequality
  • Basic knowledge of algebraic manipulation
  • Concept of ordering in real numbers
NEXT STEPS
  • Study the AM-GM inequality and its applications in proofs
  • Explore advanced topics in inequalities, such as Cauchy-Schwarz and Jensen's inequality
  • Investigate properties of exponential functions in mathematical analysis
  • Practice proving inequalities with varying conditions on variables
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Mathematicians, students studying inequalities, and anyone interested in advanced algebraic proofs will benefit from this discussion.

Albert1
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given :
$a>b>c>0$
prove :
$a^{2a}\times b^{2b}\times c^{2c}>a^{b+c}\times b^{c+a}\times c^{a+b}$
 
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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 :
$\dfrac {A}{B}=a^{(a-b)}a^{(a-c)}b^{(b-a)}b^{(b-c)}c^{(c-a)}c^{(c-b)}$
$=(\dfrac{a}{b})^{a-b}(\dfrac{b}{c})^{b-c}(\dfrac{a}{c})^{a-c}>1^0\times1^0\times 1^0=1$
 
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