Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$

Question

Reducing this whole expression i finally came to this $$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$ Here I am stuck. I can't prove this.

So I thought maybe I should try in another way. Let $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$ Hence the whole expression comes to this inequality \begin{align*} (a+b+c)^2&(a+b)(b+c)(c+a) \\ & \geq 4(ab+bc+ca)(ab^2+a^2b+bc^2+b^2c+ca^2+c^2a)\\ \implies (a+b+c)^2&\left((a+b+c)(ab+bc+ca)-abc\right)\\ & \geq 4(ab+bc+ca)\left((a+b+c)(ab+bc+ca)-3abc\right)\\ \implies (3u)^2\left(3u\times3v^2-w^3\right)&\geq 4\times3v^2\left(3u\times3v^2-3w^3\right)\\ \implies 9u^2(9uv^2-w^3)&\geq 12v^2(9uv^2-3w^3)\\ \implies 9u^3v^2-u^2w^3&\geq 12uv^4-4v^2w^3\end{align*}

Here again, I am stuck. How can I prove this inequality?

Answer 1

Another way.

Afte using your $uvw$'s substitution we see that our inequality is a linear inequality of $w^3$,

which by $uvw$ says that it's enough to prove our inequality in the following cases.

1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$ and $b=1$.

We obtain: $$a(a+1)(a-1)^2\geq0;$$ 2. Two variables are equal.

Let $b=c=1$.

We obtain: $$a^2(a-1)^2\geq0.$$ About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Answer 2

We need to prove that: $$(a+b+c)^2\prod_{cyc}(a+b)\geq4(ab+ac+bc)\sum_{cyc}(a^2b+a^2c)$$ or $$\sum_{sym}(a^4b-a^3b^2-a^3bc+a^2b^2c)\geq0.$$ Now, let $a\geq b\geq c$.

Thus, $$\sum_{sym}(a^4b-a^3b^2-a^3bc+a^2b^2c)=\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4-abc(a^2-2ab+b^2))=$$ $$=\sum_{cyc}(a-b)^2(ab(a+b)-abc)=\sum_{cyc}(a-b)^2ab(a+b-c)\geq$$ $$\geq(a-c)^2ac(a+c-b)+(b-c)^2bc(b+c-a)\geq$$ $$\geq(b-c)^2ac(a-b)+(b-c)^2bc(b-a)=(a-b)^2(b-c)^2c\geq0.$$