Proof of Inequality: Induction & Derivation Help

[PhysicsRock]

Homework Statement

Given $x_1, x_2...x_n$ and $x_i \geq 1$, prove $\prod_{i=1}^{n} (1+x_i) \geq \frac{2^n}{n+1} \left( 1 + \sum_{i=1}^{n} x_i \right)$.

Relevant Equations

None.

The assignment says proof by induction is possible, I cannot figure out how this is supposed to work out. Does somebody know the name of this by any chance? Seeing a derivation might help come up with an idea for a proof. Thank you everybody.

 

[haider]

Can you think of the steps involved in proving something by induction and list them out? That should help you realize what you have to do.

 

[PeroK]

Homework Statement:: Given $x_1, x_2...x_n$ and $x_i \geq 1$, prove $\prod_{i=1}^{n} (1+x_i) \geq \frac{2^n}{n+1} \left( 1 + \sum_{i=1}^{n} x_i \right)$.
Relevant Equations:: None.

The assignment says proof by induction is possible, I cannot figure out how this is supposed to work out. Does somebody know the name of this by any chance? Seeing a derivation might help come up with an idea for a proof. Thank you everybody.

The inductive step looks tricky. What about using some calculus?

 

[PeroK]

Actually, it doesn't need calculus. I just hacked away algebraically until it cracked!

 

[PhysicsRock]

Actually, it doesn't need calculus. I just hacked away algebraically until it cracked!

Yeah, I think I found a proof too. Purely algebraically.

 

[PeroK]

Hint: for any $x_1, \dots x_n$, transform the required inequality into an inequality in a function of $x = x_{n+1}$, where $S = \sum_{i=1}^{n} x_i \ge n$.

 

[PeroK]

Yeah, I think I found a proof too. Purely algebraically.

Using induction?

 

[PhysicsRock]

Using induction?

Partly, yes. I first ignored the $\frac{2^n}{N+1}$ and then showed that it fulfills the requirement for the inequality to hold.

 

[PeroK]

Partly, yes. I first ignored the $\frac{2^n}{N+1}$ and then showed that it fulfills the requirement for the inequality to hold.

I might post my solution when I get on my laptop later.

 

[PhysicsRock]

I might post my solution when I get on my laptop later.

Would be nice to see an alternative. Always like to see different proofs, so it would be highly appreciated :)

 

[PeroK]

To take the inductive step, we have to show that for all $x_1, \dots x_{n+1} \ge 1$:
$$\prod_{i = 1}^{n+1} (1 + x_i) \ge \frac{2^{n+1}}{n+2}\big ( 1 + \sum_{i = 1}^{n+1} x_i \big )$$Given the inductive hypothesis that this inequality holds for all $x_1, \dots x_{n} \ge 1$. Using this, we have:
$$\prod_{i = 1}^{n+1} (1 + x_i) \ge (1 + x_{n+1}) \frac{2^n}{n+1}\big ( 1 + \sum_{i = 1}^{n} x_i \big )$$We let $x = x_{n+1}$ and $S = \sum_{i = 1}^{n} x_i$. Note that $x \ge 1$ and $S \ge n$. Then it is enough to show (this is a standard phrase in proofs) that:
$$(1 + x)\frac{2^n}{n+1}\big ( 1 + S \big ) \ge \frac{2^{n+1}}{n+2}\big ( 1 + x + S \big )$$Or:
$$(n+2)(1 +x)(1 + S) - 2(n + 1)(1 + x + S) \ge 0$$Or:
$$x\big [(n+2)S - n\big ] - n(1 + S) \ge 0$$Finally, as $x \ge 1$:
$$x\big [(n+2)S - n\big ] - n(1 + S) \ge (n+2)S - n - n(1+S) = 2(S - n) \ge 0$$QED

Note that equality holds if and only if $\forall i: x_i = 1$.

 

[PeroK]

PS On a technical note, we could now turn this round and starting with $S \ge 1$ do a sequence of implications from bottom to top. But, I think it's perhaps better to leave it as a sequence of reverse implications or "enough to shows".

 

[WWGD]

This seems vaguely-related to the comparison of different types of means. Here arithmetic, geometric. But not 100% clear how, obviously .