<s>[INST] $Q\le \prod \frac{5+2x}{1+x}\le P$ find $P,Q$ 
if $x,y,z,$ are positives and $x+y+z=1$ and $$Q\le \prod_{cyc} \frac{5+2x}{1+x}\le P$$ find  maximum value of $Q$ and minimum value of $P$

This is actually a question made up myself ,so i don,t know whether a nice solution exists!?.
Finding P seems easier:AM-GM results $$\prod \frac{5+2x}{1+x}\le \frac{{\left(\sum_{cyc} \frac{5+2x}{1+x} \right)}^3}{27}$$
we write $$\sum_{cyc} \frac{5+2x}{1+x}=6+\sum \frac{3}{1+x}$$
But neither Jensen nor the tangent line method help as sign of inequality is reversed.
In fact i am more interested in finding $Q$.
I am looking for  a solution that avoids using computational aid(SOS) or ,uvw.
 [/INST] Let $x=y=z=\frac 13$ then $Q \leqslant \frac{4913}{64}.$ We will show that it's a maximum value, or
$$(5x+5y+7z)(5z+5x+7y)(5y+5z+7x) \geqslant \frac{4913}{64}(2x+y+z)(2y+z+x)(2z+x+y).$$
Let
$$\left\{\begin{aligned} & a = 2x+y+z\\& b = 2y+z+x \\& c = 2z+x+y\end{aligned}\right. \Rightarrow \left\{\begin{aligned} & x = \frac{3a-b-c}{4} \\& y = \frac{3b-c-a}{4} \\& z = \frac{3c-a-b}{4}\end{aligned}\right.$$
The inequality become
$$(11c+3b+3a)(3c+11b+3a)(3c+3b+11a) \geqslant 4913abc.$$
Using the AM-GM inequality, we have
$$(11a+3b+3c)(11b+3c+3a)(11c+3a+3b) \geqslant 17^3 \cdot \sqrt[17]{a^{11}b^3c^3} \cdot \sqrt[17]{b^{11}c^3a^3} \cdot \sqrt[17]{c^{11}a^3b^3} $$
$$= 4913abc.$$
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