$Q\le \prod \frac{5+2x}{1+x}\le P$ find $P,Q$

Question

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.

Answer 1

For $x=y=z=\frac{1}{3}$ we obtain a value $\frac{4913}{64}.$

We'll prove that it's a minimal value.

Indeed, after homogenization we need to prove that $$\prod_{cyc}\frac{7x+5y+5z}{2x+y+z}\geq\frac{4913}{64}$$ or $$\sum_{sym}(687x^3+489x^2y-1176xyz)\geq0,$$ which is true by AM-GM or by Muirhead.

Also, for $y=z\rightarrow0^+$ we obtain a value $\frac{175}{2}.$

We'll prove that it's a supremum of the expression.

Indeed, we need to prove that: $$\prod_{cyc}\frac{7x+5y+5z}{2x+y+z}\leq\frac{175}{2}$$ or $$\sum_{sym}(135x^2y+94xyz)\geq0,$$ which is obvious.

Answer 2

Finding Q via Jensen's:
Let $f(x) = \frac{ 5 + 2x } { 1 + x } $.
Let $ g(x) = \ln \frac{ 5+2x}{1+x} $, then $ g'(x) = - \frac{3}{ 2x^2 + 7x + 5 } $ and $ g'' (x) = \frac{ 3 ( 4x + 7 ) } { (2x^2 + 7x + 5 )^2 }$.
When $ x \geq 0$, $g''(x) \geq 0$, so we may apply Jensens to conclude that

$$ \sum g(x) \geq 3 g ( \frac{1}{3} ).$$

Taking $e$ to the power of both sides, we conclude that

$$ \prod f(x) \geq f(\frac{1}{3} ) ^ 3. $$

Note: The tangent line approach is similar on $g(x)$, and it works because of the condition $ g''(x) \geq 0$.

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Finding P via Step-wise smoothing
A good guess is that the maximum occurs at the extreme end (which isn't necessarily the case), which others have shown. As such, step-wise smoothing is often good approach.

Claim: When $a,b \geq 0$, $f(a) f(b) \leq f(0 ) f( a+b)$.
This follow by cross multiplying to get $ 6x^2 + 21 xy + 6y^2 \geq 0 $ which is obviously true.

Hence, $f(x)f(y) f(z) \leq f(0)f(x+y)f(z) \leq f(0) f(0) f(x+y+z) = \frac{175}{2}$.
It remains to show that this is indeed the supremum, which you can verify with $ x = y \rightarrow 0^+$.

Note: Karamata's inequality (which is a generalization of Jensen's) also gives this result directly. In fact, the above approach is a specific case of Karamata (but doesn't require as much power).

Answer 3

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 $$(11a+3b+3c)(11b+3c+3a)(11c+3a+3b) \geqslant 17^3 \cdot \sqrt[17]{a^{11}b^3c^3} \cdot \sqrt[17]{b^{11}c^3a^3} \cdot \sqrt[17]{c^{11}a^3b^3} $$ $$= 4913abc.$$