Clearing the fractions, it suffices to prove that $3\sum_{sym} a^2b + 4 \sum_{cyc} a^2b^2c \ge 30abc$. Using $\tfrac{a^2+b^2+c^2}{3} = 1$, we can homogenize the inequality to obtain the equivalent \[(a^2+b^2+c^2)\sum_{sym} a^2b + 4abc \sum_{cyc} ab \ge 10(a^2+b^2+c^2)abc .\]Observe that since $\sum_{cyc} a^2 \ge \sum_{cyc} ab$, it suffices to prove that \[\sum_{sym} a^2b + 4abc \ge 10abc.\]This is clearly true by Muirhead, so we are done. Equality holds iff $a=b=c=1$. $\blacksquare$

This is just the Cauchy-Shwartz inequality. $$\left( \frac{a^2+b^2}{2ab}+\frac{b^2+c^2}{2bc}+\frac{c^2+a^2}{2ca}\right)(ab+bc+ca)\geq(a^2+b^2+c^2)^2$$$$\frac{a^2+b^2}{2ab}+\frac{b^2+c^2}{2bc}+\frac{c^2+a^2}{2ca}+\frac{2(ab_bc+ca)}{3}\geq \frac{9}{ab+bc+ca}+\frac{2(ab+bc+ca)}{3}$$Since $ab+bc+ca\leq 3$ and $f(x)=\frac{9}{x}+\frac{2x}{3}$ is decreasing on $[0,3]$ the result follows.

Let $a, b,$ and $c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that $$\frac{a^2 + b^2}{ab} + \frac{b^2 + c^2}{bc} + \frac{c^2 + a^2}{ca} + \frac{5(ab + bc + ca)}{3} \ge 11 $$

Let $a, b,$ and $c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that $$\frac{a^2 + b^2}{ab} + \frac{b^2 + c^2}{bc} + \frac{c^2 + a^2}{ca} + \frac{4\sqrt{2}(ab + bc + ca)}{3} \ge 6+ 4\sqrt{2} $$

mudok wrote: Let $a, b,$ and $c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that $$\frac{a^2 + b^2}{ab} + \frac{b^2 + c^2}{bc} + \frac{c^2 + a^2}{ca} + \frac{4\sqrt{2}(ab + bc + ca)}{3} \ge 6+ 4\sqrt{2} $$ $\bullet$ Assuming that : $a \ge b,c .$ We need prove : $$ \frac{(a+b+c)(ab+bc+ca)-9abc}{abc}-\frac{4\sqrt{2}(a^2+b^2+c^2-ab-bc-ca)}{a^2+b^2+c^2}\ge 0$$$$\iff \frac{2a(b-c)^2+(b+c)(a-b)(a-c)}{abc}-\frac{4\sqrt{2}\left[(b-c)^2+(a-b)(a-c)\right]}{a^2+b^2+c^2}\ge 0$$$$ \iff M(b-c)^2+N(a-b)(a-c)\ge 0$$ $\bullet M=\frac{2a^2+2b^2+2c^2-4\sqrt{2}bc}{abc(a^2+b^2+c^2)}\ge \frac{3b^2+3c^2-4\sqrt{2}bc}{abc(a^2+b^2+c^2)} \ge 0$ $\bullet N=\frac{(a^2+b^2+c^2)(b+c)-4\sqrt{2}abc}{abc(a^2+b^2+c^2)}\ge \frac{\left(a^2+\dfrac{(b+c)^2}{2}\right)(b+c)-\sqrt{2}a(b+c)^2}{abc(a^2+b^2+c^2)}=\frac{(b+c)\left(a-\dfrac{b+c}{\sqrt{2}}\right)^2}{abc(a^2+b^2+c^2)} \ge 0$ .

Alternative. Adding three to both sides and using \[ \frac{a^2+b^2}{2ab} + 1 = \frac{(a+b)^2}{2ab} \]it suffices to prove \[ \sum \frac{(a+b)^2}{2ab} + \frac{2ab+2bc+2ca}{3}\ge 8. \]By Cauchy-Schwarz, we have \[ \left(\sum 2ab\right)\left(\sum \frac{(a+b)^2}{2ab}\right)\ge (2a+2b+2c)^2 \Rightarrow \sum \frac{(a+b)^2}{2ab}\ge \frac{2(a+b+c)^2}{ab+bc+ca}. \]Setting finally $t:=ab+bc+ca$ and using $3=\sum a^2$, this is equivalent to proving \[ \frac{(a+b+c)^2}{ab+bc+ca} + \frac{ab+bc+ca}{3} = \frac{3+2t}{t} + \frac{t}{3} = 2 + \frac{t}{3}+\frac{3}{t}\ge 4, \]which is obvious since $t/3 + 3/t\ge 2$. Equality holds iff $t=3$, yielding $a=b=c=1$.

mudok wrote: Let $a, b,$ and $c$ be positive real numbers such that $a^2 + b^2 + c^2 = 3$. Prove that $$\frac{a^2 + b^2}{ab} + \frac{b^2 + c^2}{bc} + \frac{c^2 + a^2}{ca} + \frac{4\sqrt{2}(ab + bc + ca)}{3} \ge 6+ 4\sqrt{2} $$ Nice.