Can This Product Inequality Be Proven for Positive x and Natural n?

[Albert1]

Given:

$$x>0,\, n\in\mathbb{N}$$

Prove:

$$ (1+x)\times\left(1+x^2 \right)\times\left(1+x^3 \right)\times\cdots\times\left(1+x^n \right)\geq\left(1+x^{\large{\frac{n+1}{2}}} \right)^n$$

 

[chisigma]

Given:

$$x>0,\, n\in\mathbb{N}$$

Prove:

$$ (1+x)\times\left(1+x^2 \right)\times\left(1+x^3 \right)\times\cdots\times\left(1+x^n \right)\geq\left(1+x^{\large{\frac{n+1}{2}}} \right)^n$$

Let's define...

$\displaystyle \alpha_{n} (x) = \prod_{k=1}^{n} (1 + x^{k})$

$\displaystyle \beta_{n} (x) = (1 + x^{\frac{n+1}{2}})^{n}\ (1)$

For x=1 is $\displaystyle \alpha_{n} = \beta_{n} = 2^{n}$, so that we analyse separately the two cases $\displaystyle 0<x<1$ and $\displaystyle x>1$. If $\displaystyle x>1$ we observe that $\alpha_{n}$ obeys to the difference equation...

$\displaystyle \alpha_{n+1} (x) = \alpha_{n}(x)\ (1+x^{n+1}),\ \alpha_{1}(x)= 1+x\ (2)$

... and if we call $\gamma_{n} (x)$ the sequence that obeys to the difference equation...

$\displaystyle \gamma_{n+1} (x) = \gamma_{n}(x)\ (1+x^{\frac{n+1}{2}}),\ \gamma_{1}(x)= 1+x\ (3)$

... we conclude that for $\displaystyle x>1$ is...

$\displaystyle \alpha_{n}(x) \ge \gamma_{n}(x) \ge \beta_{n}(x)\ (4)$

The case $\displaystyle 0 < x < 1$ will be analysed in next post...

Kind regards

$\chi$ $\sigma$

 

[kaliprasad]

we have x^a + x^b >= 2x^(a+b)/2 by AM GM inequality

adding 1 + x^(a+b) on both sides we get

(1+x^a)(1+x^b) >= 1 + 2x^(a+b)/2 + x^(a+b) = ( 1+ x^(a+b)/2)^2

Putting b = n+ 1 -a we get

(1+x^a)(1+x^(n+1- a) >= (1+ x^(n+1)/2)^2

For n even taking a from 1 to n/2 we get n/2 expressions and multiplying them out we get

(1+x)(1+x^2)( 1 + x^3) .. (1+x^n) >= (1+ x^(n+1)/2)^n as

For n odd we have n-1 ( running a from 1 to (n-1)/2 we get (n-1)/2 pairs and

As (1+x^(n+1)/2)= (1+x^(n+1)/2) and multiplying we get the result

 

[Opalg]

Given:

$$x>0,\, n\in\mathbb{N}$$

Prove:

$$ (1+x)\times\left(1+x^2 \right)\times\left(1+x^3 \right)\times\cdots\times\left(1+x^n \right)\geq\left(1+x^{\large{\frac{n+1}{2}}} \right)^n$$

Since $x^k + x^{n+1-k} - 2x^{(n+1)/2} = \bigl(x^{k/2} - x^{(n+1-k)/2}\bigr)^2 \geqslant0$, it follows that $x^k+ x^{n+1-k} \geqslant 2x^{(n+1)/2}.$ Add $1 + x^{n+1}$ to each side to see that $$(1+x^k)(1+x^{n+1-k}) = 1 + x^k+ x^{n+1-k} + x^{n+1} \geqslant 1 + 2x^{(n+1)/2} + x^{n+1} = \bigl(1 + x^{(n+1)/2}\bigr)^2.$$ If $n$ is even, multiply together the inequalities $(1+x^k)(1+x^{n+1-k}) \geqslant \bigl(1 + x^{(n+1)/2}\bigr)^2$ for $k=1,2,\ldots,n/2$ to get the result. If $n$ is odd, multiply the inequalities for $k=1,2,\ldots,(n-1)/2$, and then multiply each side by a further factor $(1 + x^{(n+1)/2})$.Edit. http://www.mathhelpboards.com/members/kaliprasad/ got there first.