Prove this inequality $\sum _{cyc}a^3+4\left(\sum _{cyc}\frac{ab}{a^2+b^2}\right)\ge 9$

Question

Let $a>0$, $b>0$ and $c>0$ such that $abc=1$. Prove that: $$a^3+b^3+c^3+4\left(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\right)\ge 9.$$

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$$L.H.S=a^3+b^3+c^3+4\left(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\right)\ge 9$$

$$\ge 3+4\left(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\right)$$

Need to prove $$3+4\left(\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\right)\ge 9$$

$$\Leftrightarrow \frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\ge \frac{3}{2}$$

This inequality is homogeneous, we assume that $a^2+b^2+c^2=3$, and need to prove $$\frac{1}{3c-c^3}+\frac{1}{3a-a^3}+\frac{1}{3b-b^3}\ge \frac{3}{2}$$

We have inequality $$\frac{1}{3a-a^3}\ge \frac{3}{8}a^2+\frac{1}{8}(*)$$

$$\Rightarrow LHS\ge \frac{3}{8}\left(a^2+b^2+c^2\right)+\frac{1}{8}.3=\frac{3}{2}$$

How to prove (*) ? $\Leftrightarrow \frac{\left(a-1\right)\left(3a^4+3a^3-5a^2-5a-8\right)}{8\left(3a-a^3\right)}\ge 0$ Right ?

Answer 1

Your way is wrong because the inequality $\sum\limits_{cyc}\frac{ab}{a^2+b^2}\geq\frac{3}{2}$ is wrong.

Try $b=c\rightarrow0^+$.

SOS helps.

Let $a\geq b\geq c$.

Hence, by AM-GM we obtain: $$b^2\sum_{cyc}\left(a^3+\frac{4ab}{a^2+b^2}-3\right)=b^2\left(\sum_{cyc}(a^3-abc)-2\sum_{cyc}\left(1-\frac{2ab}{a^2+b^2}\right)\right)=$$ $$=b^2\sum_{cyc}(a-b)^2\left(\frac{a+b+c}{2}-\frac{2}{a^2+b^2}\right)=\frac{b^2}{2}\sum_{cyc}\frac{(a-b)^2((a+b+c)(a^2+b^2)-4abc)}{a^2+b^2}\geq$$ $$\geq\frac{b^2}{2}\sum_{cyc}\frac{(a-b)^2((a+b+c)2ab-4abc)}{a^2+b^2}=b^2\sum_{cyc}\frac{(a-b)^2ab(a+b-c)}{a^2+b^2}\geq$$ $$\geq \frac{b^2(a-c)^2ac(a+c-b)}{a^2+c^2}+\frac{b^2(b-c)^2bc(b+c-a)}{b^2+c^2}\geq$$ $$\geq\frac{a^2(b-c)^2ac(a-b)}{a^2+c^2}+\frac{b^2(b-c)^2bc(b-a)}{b^2+c^2}=$$ $$=\frac{abc^3(a+b)(b-c)^2(a-b)^2}{(a^2+c^2)(b^2+c^2)}\geq0.$$ Done!

Answer 2

We can factorize \begin{align} a^3+b^3+c^3-3abc &= (a+b+c)(a^2+b^2+c^2-ab-ac-bc) \\ &= \frac{a+b+c}{2}[(a-b)^2+(a-c)^2+(b-c)^2] \end{align} and write $$\frac{2ab}{a^2+b^2} = 1-\frac{(a-b)^2}{a^2+b^2} $$ (similarly with the other terms), to rewrite, with $abc=1$, \begin{align} & a^3+b^3+c^3+4\left(\frac{ab}{a^2+b^2}+\frac{ac}{a^2+c^2}+\frac{bc}{b^2+c^2}\right) \\ &=(a^3+b^3+c^3-3abc) + 2\left[\left(\frac{2ab}{a^2+b^2}-1\right)+\left(\frac{2ac}{a^2+c^2}-1\right)+\left(\frac{2bc}{b^2+c^2}-1\right)\right]\\ &+3abc+2+2+2 \\ &=\frac{a+b+c}{2}[(a-b)^2+(a-c)^2+(b-c)^2]-2\left(\frac{(a-b)^2}{a^2+b^2}+\frac{(a-c)^2}{a^2+c^2}+\frac{(b-c)^2}{b^2+c^2}\right)+9. \end{align} It thus suffices to show that \begin{align}&\frac{a+b+c}{2}[(a-b)^2+(a-c)^2+(b-c)^2]\\ &\ge\frac{2}{a^2+b^2}(a-b)^2+\frac{2}{a^2+c^2}(a-c)^2+\frac{2}{b^2+c^2}(b-c)^2 \end{align} when $abc=1$.

Without loss of generality, suppose $a\ge b\ge c$. Then $(a-c)^2\ge (b-c)^2$, while $\frac{2}{a^2+c^2}\le\frac{2}{b^2+c^2}$. By the rearrangement inequality, we have $$\frac{2}{a^2+c^2}(a-c)^2+\frac{2}{b^2+c^2}(b-c)^2\le\frac{2}{b^2+c^2}(a-c)^2+\frac{2}{a^2+c^2}(b-c)^2.$$ Thus $\frac{2}{a^2+c^2}(a-c)^2+\frac{2}{b^2+c^2}(b-c)^2$ is also less than or equal to its average with $\frac{2}{b^2+c^2}(a-c)^2+\frac{2}{a^2+c^2}(b-c)^2$, i.e. $$\frac{2}{a^2+c^2}(a-c)^2+\frac{2}{b^2+c^2}(b-c)^2\le\left(\frac{1}{a^2+c^2}+\frac{1}{b^2+c^2}\right)[(a-c)^2+(b-c)^2].$$ We have, by AM-GM, $$a^2+c^2\ge 2ac = \frac{2}{b},$$ and similarly $b^2+c^2\ge\frac{2}{a}$, so $\frac{1}{a^2+c^2}+\frac{1}{b^2+c^2}\le \frac{b}{2}+\frac{a}{2}\le\frac{a+b+c}{2}$. It follows that \begin{align}\frac{a+b+c}{2}[(a-c)^2+(b-c)^2]&\ge\left(\frac{1}{a^2+c^2}+\frac{1}{b^2+c^2}\right)[(a-c)^2+(b-c)^2]\\ &\ge\frac{2}{a^2+c^2}(a-c)^2+\frac{2}{b^2+c^2}(b-c)^2. \end{align} Similarly, $\frac{2}{a^2+b^2}\le c$, which is certainly less than $\frac{a+b+c}{2}$ since in fact $c\le\frac{a+b+c}{3}$, so $$\frac{a+b+c}{2}(a-b)^2\ge\frac{2}{a^2+b^2}(a-b)^2.$$ Adding the two inequalities gives the desired inequality.