For three positive real numbers $a, b$ and $ c$ prove that
$$ \frac{a^3+b^3+c^3}{3} \ge \frac{a+b+c}{3} \times \frac{a^2+b^2+c^2}{3} \ge \frac{a^2b + b^2c + c^2a}{3}$$

Given Hint: Make use of Weighted AM-GM inequality

Question

For three positive real numbers $a, b$ and $ c$ prove that 
$$ \frac{a^3+b^3+c^3}{3} \ge \frac{a+b+c}{3} \times  \frac{a^2+b^2+c^2}{3} \ge \frac{a^2b + b^2c + c^2a}{3}$$

Given Hint: Make use of Weighted AM-GM inequality

abb0t · Answer

damit Wolfram!

experimentx · Answer

apparently wolfram cannot evaluate it ... since this is does not look bound. guess we cannot use method of Lagrange multiplies.

abb0t · Answer

OMG. I would of never even thought to use lagrange. Lol. I have no idea how to solve this. What math is this for?

experimentx · Answer

this should be high school level mathematics ...

abb0t · Answer

um, cauchy inequal? Lol

experimentx · Answer

looks like it ... but it was under Weighted AM-GM inequality https://www.google.com/search?q=basics+of+math+olympiad+inequalities+filetype%3Apdf&rlz=1C1ASUT_enNP447NP447&aq=f&oq=basics+of+math+olympiad+inequalities+filetype%3Apdf&aqs=chrome.0.57.8831&sourceid=chrome&ie=UTF-8

abb0t · Answer

y not multiply by 3 to get a non-fraction? Idk. I'm no math guru. Lol

experimentx · Answer

i don't know ... this one is pretty weird.

shubhamsrg · Answer

is that (a+b+c)/c ? Or you meant 3 ?

experimentx · Answer

3