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
$$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}&lt;6$$

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 $f''(a)&gt;0$. 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 $6$
${a,b,c} \geq \frac{1}{3}$, where $a^2+b^2+c^2 = a+b+c$
proof that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} &lt; 6$$
say that $$m = \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} $$
therefore $$a{b}{c}{m} = a{b}{c}(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a})  $$
$${a}{b}{c}{m} = a^3c+b^3a+c^3b$$
use the fact that $a^2+b^2+c^2 = a+b+c$ , to depress the expression by eliminating $c$
$$a^3c+b^3a+c^3b - {a}{b}{c}{m} = 0$$
$$a^2+b^2+c^2-a-b-c = 0$$
so that we would take resultant of the two polynomials
$$\operatorname{Resultant} ( a^3c+b^3a+c^3b - {a}{b}{c}{m} , a^2+b^2+c^2-a-b-c , c )$$
the results is
$$ a^2b^4m^2-a^2b^3m^2+a^4b^2m^2-a^3b^2m^2+2ab^6m-4ab^5m+4a^3b^4m-5a^2b^4m+ab^4m-2a^4b^3m-4a^3b^3m+4a^2b^3m+ab^3m+2a^5b^2m-2a^4b^2m+a^3b^2m+a^2b^2m-2a^6bm+2a^5bm+b^8-3b^7+4a^2b^6-6ab^6+3b^6-2a^3b^5-6a^2b^5+9ab^5-b^5+3a^4b^4-5a^3b^4+9a^2b^4-2ab^4-4a^5b^3+2*a^4b^3+5a^3b^3-3a^2b^3+2a^6b^2+a^5b^2-a^4b^2-2a^3b^2-2a^7b+3a^6b-a^5b-a^4b+a^8-a^7 = 0$$
so where going to prove that if ${a,b} \geq \frac{1}{3}$, then $m &lt; 6$, the maximum value of $m$ for whatsoever is always less than $6$
$$(a^2b^4-a^2b^3+a^4b^2-a^3b^2)m^2+(2ab^6-4ab^5+4a^3b^4-5a^2b^4+ab^4-2a^4b^3-4a^3b^3+4a^2b^3+ab^3+2a^5b^2-2a^4b^2+a^3b^2+a^2b^2-2a^6b+2a^5b)m+b^8-3b^7+4a^2b^6-6ab^6+3b^6-2a^3b^5-6a^2b^5+9ab^5-b^5+3a^4b^4-5a^3b^4+9a^2b^4-2ab^4-4a^5b^3+2a^4b^3+5a^3b^3-3a^2b^3+2a^6b^2+a^5b^2-a^4b^2-2a^3b^2-2a^7*b+3a^6b-a^5b-a^4b+a^8-a^7 = 0$$
but there's a problem, because of the appearance of your condition $a^2+b^2+c^2 = a+b+c$ , $m$ sometimes has a complex value, though which was not part of your statement

true test
if $a= 0.1, b= 0.2$, then $[m = 0.2857951031446771,m = 14.65420489685532]$
if $a= 0.4, b= 0.5$, then $[m = -0.04957750385466814,m = 5.129373422222015]$

the discriminate of the quadratic in $m$ gives us a condition, such that we must let the roots of $m$ to be real
$$\Delta = -a^2b^4(4b^2-4b+4a^2-4a-1)(ab^2-b^2+b-a^2+a)^2$$
$\Delta ≥ 0$ only if $-4b^2+4b-4a^2+4a+1 ≥ 0$
$$-4(b^2-b)-4(a^2-a)+1 ≥ 0$$
$$-4b(b-1)-4a(a-1)+1 ≥ 0$$
therefore $${a,b} ≤ 1$$, then $$\frac{1}{3} ≤ {a,b} ≤ 1$$
according to $$m = \frac{a^3c+b^3a+c^3b }{a{b}{c}}$$, the maximum value of $m$ happens when {a,b,c} are minimum, so setting $a=\frac{1}{3}$, $b=\frac{1}{3}$ and solving the quadratic in $m$
$$(a^2b^4-a^2b^3+a^4b^2-a^3b^2)m^2+(2ab^6-4ab^5+4a^3b^4-5a^2b^4+ab^4-2a^4b^3-4a^3b^3+4a^2b^3+ab^3+2a^5b^2-2a^4b^2+a^3b^2+a^2b^2-2a^6b+2a^5b)m+b^8-3b^7+4a^2b^6-6ab^6+3b^6-2a^3b^5-6a^2b^5+9ab^5-b^5+3a^4b^4-5a^3b^4+9a^2b^4-2ab^4-4a^5b^3+2a^4b^3+5a^3b^3-3a^2b^3+2a^6b^2+a^5b^2-a^4b^2-2a^3b^2-2a^7*b+3a^6b-a^5b-a^4b+a^8-a^7 = 0$$
gives us $m = \frac{1}{3}$ and $m = 5.75$
meaning $$m &lt; 6$$
$$\begin{matrix}
a&amp;b&amp;m\\
\frac{1}{3}&amp;\frac{1}{3}&amp;0.333,5.75\\
1&amp;1&amp;3\\
\frac{1}{3}&amp;1&amp;-5.126,5.181\\
1&amp;\frac{1}{3}&amp;2.441,4.502\\
\end{matrix}$$
Actually, this problem could be solved analytically or using Lagrange multipliers, but here the (Analytical solution).

show :

$\frac{a²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${&lt; 6}$

s.t:

${a, b, c ≥ ⅓}$ 

${a² + b² + c² = a + b + c}$



Changing the problem a little

${a, b, c ≥ 1}$

$\frac{a²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${&lt; 2}$



rearranging the function 

${a² = a + b + c - b² - c²}$

$\frac{a + b + c - b² - c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${&lt; 2}$

then 

$\frac{a}{b}$ + ${1}$ + $\frac{c}{b}$  - ${b}$ - $\frac{c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${&lt; 2}$


${b ≥ 1}$ ⇒ ${1-b}$ ≤ 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 $\frac{2c - c²}{a}$ ${&lt; 1}$  or  ${2c ≤ c²}$ but ${c≥1}$ 

then ${2c ≤ c²}$ for all ${c≥1}$