H hearts123 New member Joined Feb 22, 2019 Messages 19 Jun 8, 2019 #1 Hi everyone! I have a proving problem that I don't know how to do, at all. Any help is greatly appreciated!! Given a and b as positive real numbers, and a is not equal to b Prove that a^5 + b^5 > (a^3)(b^2) + (a^2)(b^3)

Hi everyone! I have a proving problem that I don't know how to do, at all. Any help is greatly appreciated!! Given a and b as positive real numbers, and a is not equal to b Prove that a^5 + b^5 > (a^3)(b^2) + (a^2)(b^3)

pka Elite Member Joined Jan 29, 2005 Messages 8,559 Jun 8, 2019 #2 hearts123 said: Given a and b as positive real numbers, and a is not equal to b Prove that \(\displaystyle a^5 + b^5 > (a^3)(b^2) + (a^2)(b^3)\) Click to expand... \(\displaystyle \begin{align*}(a+b)^5 &=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\\ &\ge5a^4b+10a^3b^2+10a^2b^3+5ab^4 \\&\ge 10a^3b^2+10a^2b^3\\& \text{Can you finish?} \end{align*}\)

hearts123 said: Given a and b as positive real numbers, and a is not equal to b Prove that \(\displaystyle a^5 + b^5 > (a^3)(b^2) + (a^2)(b^3)\) Click to expand... \(\displaystyle \begin{align*}(a+b)^5 &=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\\ &\ge5a^4b+10a^3b^2+10a^2b^3+5ab^4 \\&\ge 10a^3b^2+10a^2b^3\\& \text{Can you finish?} \end{align*}\)