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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 $6$

zhoraster
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math110
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    The following is also true (obtained by Mathematica using Lagrange multipliers): $$\frac{a^2}b+\frac{b^2}c+\frac{c^2}a\le 6-\frac{\rho}{a b c},$$ where $\rho$ is the root of $$23988055525253185536 x^8+7704497000298332160 x^7+8194057238162458665 x^6+2668153946412058002 x^5-9159645114203751900 x^4-2417398625137389003 x^3+270169678447885764 x^2-8530510903322967 x+85638207566543$$ near $0.0284$ – Maximilian Janisch Apr 10 '20 at 17:13
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    Also, the following stronger inequality is true. $\frac{6(a^2+b^2+c^2)}{a+b+c}>\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$, where ${a,b,c}\subset[10,41].$ All these are not for contests of course. – Michael Rozenberg Apr 10 '20 at 17:17
  • If $a,b,c = \frac{1}{3}$, you don't get more than $6$.... – mathworker21 Apr 17 '20 at 03:10

2 Answers2

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${a,b,c} \geq \frac{1}{3}$, where $a^2+b^2+c^2 = a+b+c$

proof that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} < 6$$

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 $m < 6$, 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 $$m < 6$$

$$\begin{matrix} a&b&m\\ \frac{1}{3}&\frac{1}{3}&0.333,5.75\\ 1&1&3\\ \frac{1}{3}&1&-5.126,5.181\\ 1&\frac{1}{3}&2.441,4.502\\ \end{matrix}$$

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Aderinsola Joshua
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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

$\frac{a}{b}$ + ${1}$ + $\frac{c}{b}$ - ${b}$ - $\frac{c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$

  1. ${b ≥ 1}$ ⇒ ${1-b}$ ≤ 0

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

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

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

  5. $\frac{1}{b}$ ${≤1}$ and $\frac{b-b²}{ba}$ ${≤ 0}$ and $\frac{c - c²}{ ba}$ ${≤ 0}$

then we need to proof that $\frac{c²}{a}$ ${≤ 1}$

  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 ${c≥1}$