For $a,b,c>0.$ Prove$:$ $$\displaystyle \frac {a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \geqslant \frac 32 \cdot \sqrt[6]{\dfrac{ab+bc+ca}{a^2+b^2+c^2}}$$ My try. The Buffalo Way method help here$,$ but it's not the human solution.
So I'm trying to find another as follow. Let $p=a+b+c,q=ab+bc+ca,r=abc.$
Need to prove$:$ $${\frac { \left( b-c \right) \left( a-c \right) \left( a-b \right) }{ \left( a+b \right) \left( b+c \right) \left( c+a \right) }} \geqslant 3\Big(\sqrt[6]{\frac{ab+bc+ca}{a^2+b^2+c^2}} -1\Big)$$ If $(a-b)(b-c)(a-c) \geqslant 0$ then we have the result.
If $(a-b)(b-c)(a-c) \leqslant 0$ then the inequality is equivalent to$:$ $$3\Big(1-\sqrt[6]{\frac{ab+bc+ca}{a^2+b^2+c^2}} \Big) \geqslant {\frac { \left( b-c \right) \left( c-a \right) \left( a-b \right) }{ \left( a+b \right) \left( b+c \right) \left( c+a \right) }}$$ Or $$9\Big(1-\sqrt[6]{\frac{q}{p^2-2q}}\Big)^2 \geqslant \frac{-4\,{p}^{3}r+{p}^{2}{q}^{2}+18\,pqr-4\,{q}^{3}-27\,{r}^{2}}{(pq-r)^2}$$ Let $0<\frac{q}{p^2-2q}=x^6 \leqslant 1$ thus $q={\frac {{p}^{2}{x}^{6}}{2\,{x}^{6}+1}}.$ Need to prove$:$ $$f(x)=9(1-x)^2 - \left( -4\,{p}^{3}r+{\frac {{p}^{6}{x}^{12}}{ \left( 2\,{x}^{6}+1 \right) ^{2}}}+18\,{\frac {{p}^{3}{x}^{6}r}{2\,{x}^{6}+1}}-4\,{\frac {{p}^{6}{x}^{18}}{ \left( 2\,{x}^{6}+1 \right) ^{3}}}-27\,{r}^{2} \right) \left( {\frac {{p}^{3}{x}^{6}}{2\,{x}^{6}+1}}-r \right) ^{-2 } \geqslant 0$$ Then prove when $x$ increase then $f(x)$ decrease. $(1)$
Finally$,$ we prove $f(1) \geqslant 0$ or ${\frac { \left( {p}^{3}-27\,r \right) ^{2}}{ \left( {p}^{3}-3\,r \right) ^{2}}} \geqslant 0.$
However I can't not prove $(1).$ Can some one help me end it or another solution$?$
Thanks for a real lot!
See also at https://artofproblemsolving.com/community/c6h2206642p16666503
You chose a right way!
I like the $uvw$'s notation:
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, by your work we need to prove that: $$3(a+b)(a+c)(b+c)\left(1-\sqrt[6]{\frac{ab+ac+bc}{a^2+b^2+c^2}}\right)\geq(a-b)(a-c)(b-c)$$ or $$3(9uv^2-w^3)\left(1-\sqrt[6]{\frac{v^2}{3u^2-2v^2}}\right)\geq(a-b)(a-c)(b-c).$$ Now, let $\frac{v^2}{3u^2-2v^2}=t^6,$ where $t>0$.
Thus, since $u^2\geq v^2$, we obtain $0<t\leq1$, $u^2=\frac{1+2t^6}{3t^6}v^2$ and we need to prove that: $$9(9uv^2-w^3)^2(1-t)^2\geq\prod_{cyc}(a-b)^2$$ or $$(9uv^2-w^3)^2(1-t)^2\geq3(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)$$ or $$(t^2-2t+4)w^6+6u(2u^2-3(t^2-2t+2)v^2)w^3+9(9t^2-18t+8)u^2v^4+12v^6\geq0.$$ Now, since $$9(9t^2-18t+8)u^2v^4+12v^6=9(9t^2-18t+8)\cdot\frac{1+2t^6}{3t^6}v^6+12v^6=$$ $$=\frac{3v^6(18t^8-36t^7+20t^6+9t^2-18t+8)}{t^6}>0,$$ it's enough to prove our inequality for $$2u^2-3(t^2-2t+2)v^2<0$$ or $$\frac{2(1+2t^6)}{3t^6}-3(t^2-2t+2)<0$$ or $$9t^8-18t^7+14t^6-2>0,$$ which with $0<t\leq1$ gives $0.85...<t\leq1$ and it's enough to prove that $$9u^2(2u^2-3(t^2-2t+2)v^2)^2-(t^2-2t+4)(9(9t^2-18t+8)u^2v^4+12v^6)\leq0$$ or $$\frac{1+2t^6}{t^6}\left(\frac{2(1+2t^6)}{3t^6}-3(t^2-2t+2)\right)^2-(t^2-2t+4)\left(\frac{1+2t^6}{t^6}(9t^2-18t+8)+4\right)\leq0$$ or $$(t-1)^2(81t^{18}+t^{16}+2t^{15}+3t^{14}+4t^{13}+59t^{12}+6t^{11}+4t^{10}+$$ $$+2t^9-2t^7+5t^6-6t^5-5t^4-4t^3-3t^2-2t-1)\geq0,$$ which is true for $9t^8-18t^7+14t^6-2>0$ and $0<t\leq1$.