Inequality with positive real numbers a and b

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SUMMARY

The discussion centers on proving the inequality \( a^a b^b + a^b b^a \leq 1 \) for positive real numbers \( a \) and \( b \) constrained by the condition \( a + b = 1 \). The proof utilizes the properties of logarithmic functions and the application of Jensen's inequality, demonstrating that the maximum value of the expression occurs when \( a \) and \( b \) are equal. This leads to the conclusion that the inequality holds true for all positive \( a \) and \( b \) satisfying the given condition.

PREREQUISITES
  • Understanding of inequalities, particularly Jensen's inequality
  • Familiarity with properties of logarithmic functions
  • Basic knowledge of calculus and optimization techniques
  • Concept of symmetric functions in algebra
NEXT STEPS
  • Study Jensen's inequality and its applications in optimization
  • Explore the properties of logarithmic functions in depth
  • Learn about symmetric functions and their significance in inequalities
  • Investigate other inequalities involving positive real numbers, such as AM-GM inequality
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Mathematicians, students studying real analysis, and anyone interested in the properties of inequalities involving real numbers.

anemone
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Let $a$ and $b$ be positive real numbers such that $a+b=1$. Prove that $a^ab^b+a^bb^a\le 1$.
 
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We have
$1= a+ b = a^{a+b} + b^{a+b}$
So $1- (a^ab^b + a^b b^a)$
$= a^{a+b} + b^{a+b} - (a^ab^b + a^b b^a)$
$= a^a(a^b-b^b) + b^a(b^b-a^b) = (a^a - b^a)(a^b - b^b)$
For a > b both the terms are non -ve so we have and if b > a then both terms are -ve and hence above is positive

$1- (a^ab^b + a^b b^a) >=0$ and hence the result
 

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