Prove that, for any positive real numbers a, b, and c, $a^3b^2 + a^2b^3 + a^3c^2 + a^2c^3 + b^3c^2 + b^2c^3 \geq \frac{a^5 + b^5 + c^5}{5}$.

Question

For this question I assume you would start with a rewritten form so that $a^3b^2 + a^2b^3 + a^3c^2 + a^2c^3 + b^3c^2 + b^2c^3$ = $(a^3 + b^3 + c^3)(a^2 + b^2 + c^2) - (a^5 + b^5 + c^5)$, but how would I progress from there to prove the inequality? I have tried algebraic manipulation, but it has led me no where. I have also tried the AM-GM-RMS inequality, but no luck! I might have made an algebraic error somewhere. Any help is appreciated, thanks!

Answer 1

$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5$ .for easy simplification note that($(a+b)^5=\sum_{k=0}^{5}{5 \choose k}a^{5-k}b^k$)

$(a+c)^5=a^5+5a^4c+10a^3c^2+10a^2c^3+5ac^4+c^5$

$(c+b)^5=c^5+5c^4b+10c^3b^2+10c^2b^3+5cb^4+b^5$

then $(a+b)^5+(a+c)^5+(c+b)^5=2a^5+2b^5+2c^5+5ab^4+5ac^4+5cb^4+10a^2b^3+10a^2c^3+10c^2b^3+10a^3b^2+10a^3c^2+10c^3b^2+5a^4b++5a^4c+5c^4b$

and this implie that:$\frac{(a+b)^5+(a+c)^5+(c+b)^5}{10}-\frac{2a^5+2b^5+2c^5}{10}\geq a^2b^3+a^2c^3+c^2b^3+a^3b^2+a^3c^2+c^3b^2$.(beacuse if $x=y+w$ then$ x \geq y,w$ for any positive numbers $x,y,x$)

so :$-\frac{2a^5+2b^5+2c^5}{10}\geq a^2b^3+a^2c^3+c^2b^3+a^3b^2+a^3c^2+c^3b^2-\frac{(a+b)^5+(a+c)^5+(c+b)^5}{10}$

so finally ;$\frac{a^5+b^5+c^5}{5}\leq -(a^2b^3+a^2c^3+c^2b^3+a^3b^2+a^3c^2+c^3b^2)+\frac{(a+b)^5+(a+c)^5+(c+b)^5}{10}$ (**)

this inequality holds for any positive numbers, but your inequality isn't always true, and you can wonder if the left hand side of your inequality will be smaller or equal the right hand side of (**),so you will see it is not always true.