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.

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- Mar 7th 2011, 04:35 AMchris86Proving (1+x1.t)(1+x2.t)...(1+xn.t)>=(1+t)^n.
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. - Mar 7th 2011, 10:35 AMgirdav
We can show that $\displaystyle \dfrac{\left(1+\frac t{x_i}\right)(1+tx_i)}{(1+t)^2}\geq 1$ hence we have either $\displaystyle (1+tx_1)\ldots(1+tx_n)\geq (1+t)^n$ or $\displaystyle \left(1+\dfrac t{x_1}\right)\ldots \left(1+\dfrac t{x_n}\right) \geq (1+t)^n$. If it is the second case, put for $\displaystyle t\neq 0$ $\displaystyle f(t) =\left(1+\dfrac t{x_1}\right)\ldots \left(1+\dfrac t{x_n}\right)-(1+t)^n$ and compute $\displaystyle f\left(\dfrac 1t\right).$

- Mar 7th 2011, 11:44 AMemakarov
Nice. Could you explain one thing? If I understand correctly, you claim that $\displaystyle \dfrac{\prod_{i=1}^n(1+tx_i)}{(1+t)^n}\cdot\dfrac{ \prod_{i=1}^n(1+t/x_i)}{(1+t)^n}\ge1$, from where either the first or the second factor is $\displaystyle \ge 1$. Theoretically, the choice of which factor exceeds 1 can depend on $\displaystyle t$. However, it seems that later you require that either

$\displaystyle \prod_{i=1}^n(1+tx_i)\geq (1+t)^n$*for all*$\displaystyle t$ or $\displaystyle \prod_{i=1}^n(1+t/x_i)\geq (1+t)^n$*for all*$\displaystyle t$. This allows you substituting $\displaystyle 1/t$ for $\displaystyle t$ in case the second alternative is true. But is it possible, say, that the second alternative is true for $\displaystyle t = 5$, but the first alternative is true for $\displaystyle t = 1/5$? - Mar 8th 2011, 09:03 AMgirdav
You're right, what I did cannot be correct because I didn't use the fact that $\displaystyle x_1\ldots x_n=1$.

I think we can show the result by induction. It's not difficult for $\displaystyle n=2$ and it's true for $\displaystyle n$ and $\displaystyle x_1\ldots x_{n+1}=1$ were $\displaystyle x_i$ is positive.

We assume $\displaystyle x_{n+1}\leq 1$ and $\displaystyle x_n\geq 1$ (by symmetry ). Let $\displaystyle y_i=x_i$ if $\displaystyle i\leq n-1$ and $\displaystyle y_n=x_nx_{n+1}$. We have $\displaystyle \prod_{i=1}^n(1+y_it)\geq (1+t)^n $for any $\displaystyle t\geq 0$ hence $\displaystyle \prod_{i=1}^{n-1}(1+x_it)(1+x_nx_{n+1}t)\geq (1+t)^n $. Now we have to show that

$\displaystyle (1+x_nx_{n+1}t)(1+t)\leq (1+x_{n+1}t)(1+x_nt)$. If we compute the difference we get $\displaystyle (1+x_{n+1}t)(1+x_nt) -(1+x_nx_{n+1}t)(1+t)=t(x_{n+1}+x_n-x_nx_{n+1}-1)t(1-x_n)(x_{n+1}-1)$ which is positive. - Mar 10th 2011, 03:30 PMawkward
Let

$\displaystyle f(x) = \ln(1 + e^x)$.

It's easy to show that the second derivative is positive, so f is convex. By Jenson's Inequality,

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

I.e.,

$\displaystyle \ln(1 + e^{(a_1 + a_2 + \dots + a_n)/n}) \leq \ln(1 + e^{a_1})/n + \ln(1 + e^{a_2})/n + \dots + \ln(1 + e^{a_n})/n$

for any $\displaystyle a_1, a_2, \dots , a_n$.

Exponentiating,

$\displaystyle 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 $\displaystyle a_i = \ln(x_i t)$, so

$\displaystyle e^{(a_1 + a_2 + \dots + a_n)/n} = (x_1 x_2 \cdots x_n)^{1/n} t = t$

and the inequality becomes

$\displaystyle 1+t \leq (1 + x_1 t)^{1/n} (1 + x_2 t)^{1/n} \cdots (1 + x_n t)^{1/n}$.

Finally, raise both sides to the nth power.