Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$

Question

Let $a>0$,$b>0$ and $c>0$. Prove that: $$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$

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$$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$

$\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$

$\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$

$\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$

$\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$

I can't continue. Help me

Answer 1

The following quantity is clearly positive \begin{eqnarray*} \sum_{perms} a^2 b (a-b)^2 \geq 0 \end{eqnarray*} This can be rearranged to \begin{eqnarray*} 3(a^2+b^2+c^2)(a+b)(b+c)(c+a) \geq (a+b+c) \left( \sum_{cyc} (a^2 +b^2)(c+a)(c+b) \right) \end{eqnarray*} Now divide this by $(a+b+c)(a+b)(b+c)(c+a)$ and we have \begin{eqnarray*} \frac{3(a^2+b^2+c^2)}{(a+b+c)} \geq \frac{a^2 +b^2}{a+b}+ \frac{b^2 +c^2}{b+c}+ \frac{c^2 +a^2}{c+a}. \end{eqnarray*}

Answer 2

your inequality is equivalent to $$ab(a-b)(a^2-b^2)+ac(a-c)(a^2-c^2)+bc(b-c)(b^2-c^2)\geq 0$$ which is true.

Answer 3

Now, by C-S $$\sum_{cyc}\frac{1}{a+b}\geq\frac{9}{\sum\limits_{cyc}(a+b)}=\frac{9}{2(a+b+c)}.$$ Thus, it remains to prove that $$a^2+b^2+c^2+\frac{9abc}{a+b+c}\geq2(ab+ac+bc)$$ or $$\sum_{cyc}(a^3-a^2b-a^2c+abc)\geq0,$$ which is Schur.

Done!