Given three positive numbers $a, b, c$ so that $8abc\geqq a+ b+ c+ 5$. Prove that $$\frac{1}{a+ 2b}+ \frac{1}{b+ 2c}+ \frac{1}{c+ 2a}\leqq 1$$
The best $constant$ here is also $8$, which I found by using discriminant and $uvw$. On the other hand, how do I make condition homogeneous as $\lceil$https://math.stackexchange.com/a/3140810/689452$\rfloor$ ?
Let $\sum\limits_{cyc}\frac{1}{a+2b}>1$, $a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $$\sum_{cyc}\frac{1}{x+2y}=1.$$ Thus, $$\sum_{cyc}\frac{1}{kx+2ky}>1,$$ which gives $$0<k<1.$$ But the condition gives $$k(8k^2xyz-x-y-z)\geq5,$$ which says $$8k^2xyz-x-y-z>0$$ and we obtain: $$5\leq k(8k^2xyz-x-y-z)<8k^2xyz-x-y-z<8xyz-x-y-z,$$ which is a contradiction because we'll prove now that $$8xyz-x-y-z\leq5.$$ Indeed, we need to prove that $$8xyz\left(\sum\limits_{cyc}\frac{1}{x+2y}\right)^3-(x+y+z)\sum\limits_{cyc}\frac{1}{x+2y}\leq5.$$ Now, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v.$
Thus, $u\geq0$, $v\geq0$ and we need to prove that: $$21870(u^2-uv+v^2)x^7+2187(18u^3+u^2v+15uv^2+18v^3)x^6+$$ $$+162(161u^4+218u^3v+402u^2v^2+596uv^3+161v^4)x^5+$$ $$+27(278u^5+889u^4v+2345u^3v^2+4691u^2v^3+2551uv^4+278v^5)x^4+$$ $$+2(400u^6+2688u^5v+12876u^4v^2+38915u^3v^3+39660u^2v^4+9924uv^5+400v^6)x^3+$$ $$+(8u^7+236u^6v+4398u^5v^2+21367u^4v^3+38341u^3v^4+20490u^2v^5+2108uv^6+8v^6)x^2+$$ $$+4uv(-4u^6+66u^5v+639u^4v^2+1801u^3v^3+1914u^2v^4+492uv^5+8v^6)x+4u^2v^2(u+2v)^2(2u^3+19u^2v+2v^3)\geq0,$$ for which it's enough to prove that $$8u^7x^2+4uv(-4u^6)x+4u^2v^2\cdot u^2\cdot2u^3\geq0$$ or $$8u^7x^2-16u^7vx+8u^7v^2\geq0$$ or $$8u^7(x^2-2vx+v^2)\geq0$$ or $$8u^7(x-v)^2\geq0$$ and we are done!