Inequality Challenge: Prove $\sum \frac{x^3}{x^2+xy+y^2}\geq\frac{a+b+c}{3}$

[Albert1]

$a,b,c \in N$,prove :
$\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\geq\dfrac{a+b+c}{3}$

 

[PaulRS]

The result is true for $a,b,c>0$ in general.

Spoiler

By normalizing the inequality we may assume that $a+b+c=1$. $(*)$

Now rewrite the inequality we are to prove, that is
$$
\sum_{\text{cyclic}} \frac{a^3}{a^2+ab+b^2} \geq \frac{1}{3}\,,\qquad (**)
$$
as
$$
\sum_{\text{cyclic}} \frac{a}{1+(b/a)+(b/a)^2} \geq \frac{1}{3}\,.
$$

Observe that $x\mapsto 1/(1+x+x^2)$ is convex in $(0,\infty)$ because its second derivative is $\frac{6 \, {\left(x + 1\right)} x}{{\left(x^{2} + x + 1\right)}^{3}} > 0$, hence by Jensen's inequality with weights $a,b,c$ we get
$$
\sum_{\text{cyclic}} \frac{a}{1+(b/a)+(b/a)^2} \geq \frac{1}{1+\left(\sum_{\text{cyclic}}a\,(b/a)\right)+\left(\sum_{\text{cyclic}}a\,(b/a)\right)^2} = \frac{1}{1+1+1^2} =\frac{1}{3}\,.
$$

$(*)$ If $a+b+c=\lambda \neq 1$, make the change of variables $a \mapsto a/\lambda\,, b \mapsto b/\lambda \,, c \mapsto c/\lambda$ and the $\lambda$s will cancel out when considering both sides.

$(**)$ When we have variables $(a,b,c)$, the notation $\sum_{\text{cyclic}} f(a,b,c)$ simply means $f(a,b,c)+f(b,c,a)+f(c,a,b)$.

 

[Albert1]

$a,b,c \in N$,prove :
$\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\geq\dfrac{a+b+c}{3}$

indeed even $a,b,c\in R^+$,this statement is also true
hint :

Spoiler

at first prove the following statement:
$a,b,c \in R^+$
$\dfrac{a^3}{a^2+ab+b^2}\geq a-\dfrac {a+b}{3}$

 

[Albert1]

indeed even $a,b,c\in R^+$,this statement is also true
hint :

Spoiler

at first prove the following statement:
$a,b,c \in R^+$
$\dfrac{a^3}{a^2+ab+b^2}\geq a-\dfrac {a+b}{3}$

Spoiler

for $a^2+ab+b^2\geq 3ab$
$\dfrac{a^3}{a^2+ab+b^2}=\dfrac{a^3+a^2b+ab^2-ab(a+b)}{a^2+ab+b^2}=a-\dfrac {ab(a+b)}{a^2+ab+b^2}\geq a-\dfrac {a+b}{3}---(1)$
likewise :$\dfrac{b^3}{b^2+bc+c^2}\geq b-\dfrac {b+c}{3}---(2)$
$\dfrac{c^3}{c^2+ca+a^2}\geq c-\dfrac {c+a}{3}---(3)$
$(1)+(2)+(3) $ we get the reslut