Given real numbers $a,b,c >0$ and $a+b+c=3$. Prove that $\frac{a^2}{b}+\frac{b^2}{c} +\frac{c^2}{a} \ge3 + \frac{4}{3}\max\{(a-b)^2;(b-c)^2;(c-a)^2\}$ My first way is to put the inequality under the same degree, so I multipled both sides to $(a+b+c)$, however it leaded to a hard to solve result. Can anyone help me with this problem?

For the current inequality of $ \sum \frac{a^2}{b} \geq 3 + \frac{4}{3} \max ( a-b)^2 $:

*

*Replace 3 with $ a+b+c$. The reason for this choice is the "well-known" inequality: $ \sum \frac{ a^2}{b} \geq \sum a $. We're then asking how much more leeway there is in the difference of these terms.


*We can write the difference as a as Sum of Squares (Figure out how to do this before looking at the hint.). Thus, we want to show that

 $$ \sum \frac{ (a-b)^2 } { b}  \geq \frac{4}{3} \max ( a-b)^2.$$

This is just Cauchy Schwarz:

*

*Numerator: $[ \sum | a - b | ] ^2= [2 ( \max (a,b,c) - \min (a, b, c) )]^2 = 4 \max((a-b)^2)$.

*Denominator: $a + b + c  = 3$.

Equality holds in the Cauchy Schwarz iff:

*

*For $ a \geq b \geq c$: $\frac{ a-b}{b} = \frac{b-c}{c} = \frac{a-c}{a} , a+b+c = 3$ $\Rightarrow (a, b, c) = (1, 1, 1) , (\frac{3}{2} , \frac{ 3 \sqrt{5} - 3 } { 4} , \frac{ 9 - 3 \sqrt{5}}{ 4} ). $

*For $ a \geq c \geq b$: $\frac{ a-b}{b} = \frac{c-b}{c} = \frac{a-c}{a} , a+b+c =3 $, which leads to the same cases.

