How prove this inequality $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}<6$ let $a,b,c\ge\dfrac{1}{3}$,and such $$a^2+b^2+c^2=a+b+c$$
show that
$$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}<6$$
I try:$$a-a^2=(b^2-b)+(c^2-c)\ge-\dfrac{1}{4}-\dfrac{1}{4}=-\dfrac{1}{2}$$
so we have
$$a^2-a-\dfrac{1}{2}\le 0\Longrightarrow a\le\dfrac{1+\sqrt{3}}{2}$$
so we have
$$a,b,c\in [\dfrac{1}{3},\dfrac{1+\sqrt{3}}{2}]$$
let $f(a)=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}$,and $f''(a)>0$. so $LHS$ is maximum when $a,b,c=\{\dfrac{1}{3},\dfrac{1+\sqrt{3}}{2}\}$.but I found all case this value is big $6$
 A: Actually, this problem could be solved analytically or using Lagrange multipliers, but here the (Analytical solution).
show :
$\frac{a²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 6}$
s.t:
${a, b, c ≥ ⅓}$ 
${a² + b² + c² = a + b + c}$

Changing the problem a little
${a, b, c ≥ 1}$
$\frac{a²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$

rearranging the function 
${a² = a + b + c - b² - c²}$
$\frac{a + b + c - b² - c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$
then 
$\frac{a}{b}$ + ${1}$ + $\frac{c}{b}$  - ${b}$ - $\frac{c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$


*

*${b ≥ 1}$ ⇒ ${1-b}$ ≤ 0

*$\frac{c}{b}$ - $\frac{c²}{b}$ = $\frac{c-c²}{b}$   ⇒ $\frac{c-c²}{b}$ ${≤ 0}$

*$\frac{a}{b}$ + $\frac{c²}{a}$ = $\frac{a² + bc²}{bc}$ =   $\frac{(a + b + c - b² - c² ) + bc² }{ba}$

*$\frac{1}{b}$ + $\frac{b-b²}{ba}$ + $\frac{c - c²}{ba}$ + $\frac{c²}{a}$

*$\frac{1}{b}$ ${≤1}$ and $\frac{b-b²}{ba}$ ${≤ 0}$ and $\frac{c - c²}{ ba}$ ${≤ 0}$
then we need to proof that $\frac{c²}{a}$ ${≤ 1}$


*$\frac{c² + a² - a² }{a}$
= $\frac{c² + a²}{a}$ - ${a}$ 
= $\frac{c² + (a + b + c - b² - c²)}{a}$ - ${a}$
= ${1 - a}$ +  $\frac{2c - c²}{a}$ + $\frac{b-b²}{a}$
${1-a}$ ${≤ 0}$ and 
$\frac{b-b²}{a}$ ${≤ 0}$
we need to proof that $\frac{2c - c²}{a}$ ${< 1}$  or  ${2c ≤ c²}$ but ${c≥1}$ 
then ${2c ≤ c²}$ for all ${c≥1}$ 
