Problem. (Le Khanh Sy) For $a,b,c>0.$ Prove$:$ $$\sum \dfrac{a+b}{c} \geq 2\sqrt{(a+b+c)\Big(\dfrac{a}{bc} +\dfrac{b}{ca}+ \dfrac{c}{ab}\Big)}$$
My proof. After squaring ... it's
$$4\,{b}^{2}{c}^{2} \left( a-b \right) \left( a-c \right) +4\,{c}^{2}{a }^{2} \left( b-c \right) \left( b-a \right) +4\,{a}^{2}{b}^{2} \left( c-a \right) \left( c-b \right) + \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( c-a \right) ^{2} \geqslant 0,$$
which is true!
I also know another proof by AM-GM of another author$,$ we have$:$
$$\text{RHS} \leqslant \dfrac{1}{c}(a+b+c)+c\Big(\dfrac{a}{bc} +\dfrac{b}{ca}+ \dfrac{c}{ab}\Big)\leqslant \text{LHS}$$
The last inequality is$:$ $$(a-c)(c-b) \geqslant 0,$$ which is obvious if $c\equiv \text{mid}\, \{a,b,c\}.$
But is there another proof$?$
After replacing $a$ at $\frac{1}{a}$,$b$ at $\frac{1}{b}$ and $c$ at $\frac{1}{c}$ we need to prove that $$\sum_{cyc}(a^2b+a^2c)\geq2\sqrt{(ab+ac+bc)(a^2b^2+a^2c^2+b^2c^2)},$$ which is true by C-S twice: $$\sum_{cyc}(a^2b+a^2c)=(a+b+c)(a^2+b^2+c^2)-(a^3+b^3+c^3)=$$ $$=\sqrt{\sum_{cyc}(a^2+2bc)\sum_{cyc}(a^4+2a^2b^2)}-\sum_{cyc}a^3\geq$$ $$\geq\sqrt{\sum_{cyc}a^2\sum_{cyc}a^4}+2\sqrt{\sum_{cyc}ab\sum_{cyc}a^2b^2}-\sum_{cyc}a^3\geq$$ $$\geq\sum_{cyc}a^3+2\sqrt{\sum_{cyc}ab\sum_{cyc}a^2b^2}-\sum_{cyc}a^3=2\sqrt{\sum_{cyc}ab\sum_{cyc}a^2b^2}.$$