Proving (1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.

**chris86** · March 7th 2011, 04:35 AM

Let x1,x2,...,xn be positive numbers satisfying
(x1)(x2)(x3)...(xn)=1.
Show that for any t>=0,
(1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.

**girdav** · March 7th 2011, 10:35 AM

We can show that $\dfrac{\left(1+\frac t{x_i}\right)(1+tx_i)}{(1+t)^2}\geq 1$ hence we have either $(1+tx_1)\ldots(1+tx_n)\geq (1+t)^n$ or $\left(1+\dfrac t{x_1}\right)\ldots \left(1+\dfrac t{x_n}\right) \geq (1+t)^n$ . If it is the second case, put for $t\neq 0$ $f(t) =\left(1+\dfrac t{x_1}\right)\ldots \left(1+\dfrac t{x_n}\right)-(1+t)^n$ and compute $f\left(\dfrac 1t\right).$

**emakarov** · March 7th 2011, 11:44 AM

Nice. Could you explain one thing? If I understand correctly, you claim that $\dfrac{\prod_{i=1}^n(1+tx_i)}{(1+t)^n}\cdot\dfrac{ \prod_{i=1}^n(1+t/x_i)}{(1+t)^n}\ge1$ , from where either the first or the second factor is $\ge 1$ . Theoretically, the choice of which factor exceeds 1 can depend on $t$ . However, it seems that later you require that either
$\prod_{i=1}^n(1+tx_i)\geq (1+t)^n$ for all $t$ or $\prod_{i=1}^n(1+t/x_i)\geq (1+t)^n$ for all $t$ . This allows you substituting $1/t$ for $t$ in case the second alternative is true. But is it possible, say, that the second alternative is true for $t = 5$ , but the first alternative is true for $t = 1/5$ ?

**girdav** · March 8th 2011, 09:03 AM

Originally Posted by emakarov

However, it seems that later you require that either
$\prod_{i=1}^n(1+tx_i)\geq (1+t)^n$ for all $t$ or $\prod_{i=1}^n(1+t/x_i)\geq (1+t)^n$ for all $t$ . This allows you substituting $1/t$ for $t$ in case the second alternative is true. But is it possible, say, that the second alternative is true for $t = 5$ , but the first alternative is true for $t = 1/5$ ?

You're right, what I did cannot be correct because I didn't use the fact that $x_1\ldots x_n=1$ .
I think we can show the result by induction. It's not difficult for $n=2$ and it's true for $n$ and $x_1\ldots x_{n+1}=1$ were $x_i$ is positive.
We assume $x_{n+1}\leq 1$ and $x_n\geq 1$ (by symmetry ). Let $y_i=x_i$ if $i\leq n-1$ and $y_n=x_nx_{n+1}$ . We have $\prod_{i=1}^n(1+y_it)\geq (1+t)^n$ for any $t\geq 0$ hence $\prod_{i=1}^{n-1}(1+x_it)(1+x_nx_{n+1}t)\geq (1+t)^n$ . Now we have to show that
$(1+x_nx_{n+1}t)(1+t)\leq (1+x_{n+1}t)(1+x_nt)$ . If we compute the difference we get $(1+x_{n+1}t)(1+x_nt) -(1+x_nx_{n+1}t)(1+t)=t(x_{n+1}+x_n-x_nx_{n+1}-1)t(1-x_n)(x_{n+1}-1)$ which is positive.

**awkward** · March 10th 2011, 03:30 PM

Originally Posted by chris86

Let x1,x2,...,xn be positive numbers satisfying
(x1)(x2)(x3)...(xn)=1.
Show that for any t>=0,
(1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.

Let
$f(x) = \ln(1 + e^x)$ .
It's easy to show that the second derivative is positive, so f is convex. By Jenson's Inequality,

$f((a_1 + a_2 + \dots + a_n)/n) \leq f(a_1)/n + f(a_2)/n + \dots + f(a_n)/n$

I.e.,
$\ln(1 + e^{(a_1 + a_2 + \dots + a_n)/n}) \leq \ln(1 + e^{a_1})/n + \ln(1 + e^{a_2})/n + \dots + \ln(1 + e^{a_n})/n$
for any $a_1, a_2, \dots , a_n$ .

Exponentiating,
$1 + e^{(a_1 + a_2 + \dots + a_n)/n} \leq (1+ e^{a_1})^{1/n} (1+ e^{a_2})^{1/n} \cdots (1+ e^{a_n})^{1/n}$ .

Now let $a_i = \ln(x_i t)$ , so
$e^{(a_1 + a_2 + \dots + a_n)/n} = (x_1 x_2 \cdots x_n)^{1/n} t = t$
and the inequality becomes
$1+t \leq (1 + x_1 t)^{1/n} (1 + x_2 t)^{1/n} \cdots (1 + x_n t)^{1/n}$ .

Finally, raise both sides to the nth power.

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