<s>[INST] Prove that $a^2bc + ab^2c + abc^2 \le a^3b+ac^3+b^3c$ Note: $a,b$ and $c$ are positive real numbers.
I tried to use excel and I believe that, after going through a bunch of numbers, this preposition is true. However, I do not know how to prove it mathematically. Can someone help me to prove this question.
 [/INST] You have to show
$$abc(a+b+c)\leq a^3b+ac^3+b^3c$$
or (since $a,b,c > 0)$
$$a+b+c \leq \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a}$$
Now, Cauchy-Schwarz inequality helps as follows
$$(a+b+c)^2 = \left(\sqrt c\frac a{\sqrt c}+\sqrt a \frac b{\sqrt a }+\sqrt b \frac c{\sqrt b } \right)^2\stackrel{C.-S.}{\leq} (c+a+b)\cdot \left(\frac{a^2}{c}+ \frac{b^2}{a} + \frac{c^2}{b} \right)$$
Hence,
$$a+b+c \leq \frac{a^2}{c} + \frac{c^2}{b} + \frac{b^2}{a}$$
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