# Thread: Interesting problem, any proofs?

1. ## Interesting problem, any proofs?

Let $\displaystyle x_{1},x_{2},...,x_{n}$ be positive real numbers. Prove that

$\displaystyle n^2\leq(x_1+x_2+...+x_n)(\frac{1}{x_1}+\frac{1}{x_ 2}+...+\frac{1}{x_n})$ and

$\displaystyle \frac{x_1+x_2+...+x_n}{\sqrt{n}}\leq\sqrt{(x_1)^2+ (x_2)^2+...+(x_n)^2}$

I really have no idea how to even begin this... any ideas?

2. Originally Posted by paupsers
Let $\displaystyle x_{1},x_{2},...,x_{n}$ be positive real numbers. Prove that

$\displaystyle n^2\leq(x_1+x_2+...+x_n)(\frac{1}{x_1}+\frac{1}{x_ 2}+...+\frac{1}{x_n})$ and

$\displaystyle \frac{x_1+x_2+...+x_n}{\sqrt{n}}\leq\sqrt{(x_1)^2+ (x_2)^2+...+(x_n)^2}$

I really have no idea how to even begin this... any ideas?
They are both consequence of Cauchy-Schwarz inequality: $\displaystyle |\vec{x}\cdot\vec{y}|\leq \|\vec{x}\|\|\vec{y}\|$. I let you find what to take as $\displaystyle \vec{x}$ and $\displaystyle \vec{y}$.

3. I've been looking at this problem for a while, and I can't figure out how to isolate the n's on one side, and how that is going to relate to the right-hand side.

4. Originally Posted by paupsers
I've been looking at this problem for a while, and I can't figure out how to isolate the n's on one side, and how that is going to relate to the right-hand side.

The Cauchy-Schwarz inequality expressed for sums is:

$\displaystyle \left(\sum_{i=1}^n a_ib_i\right)^2\leq\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)$

Try letting $\displaystyle a_i=\sqrt{x_i}$ and $\displaystyle b_i=\frac{1}{\sqrt{x_i}}$

You can apply similar reasoning to the second problem.

5. I understand the first problem now, but I'm having a really hard time finding $\displaystyle a_i$ and $\displaystyle b_i$ for the second one. Can you explain the method of how to find them for the second one so I can use that method for the rest of the problems in my book?

6. Originally Posted by paupsers
I understand the first problem now, but I'm having a really hard time finding $\displaystyle a_i$ and $\displaystyle b_i$ for the second one. Can you explain the method of how to find them for the second one so I can use that method for the rest of the problems in my book?
Take the square root of the C-S Inequality:

$\displaystyle \sum_{i=1}^n a_ib_i\leq\sqrt{\sum_{i=1}^n a_i^2}\sqrt{\sum_{i=1}^n b_i^2}$

Now rearrange your problem a bit:

$\displaystyle (x_1+x_2+...+x_n)\leq\sqrt{(x_1)^2+(x_2)^2+...+(x_ n)^2}\sqrt{n}$

Do you see what to do now?

7. Does $\displaystyle a_i=x_1+x_2+...+x_n$ and $\displaystyle b_i=1$?

8. Originally Posted by paupsers
Does $\displaystyle a_i=x_1+x_2+...+x_n$ and $\displaystyle b_i=1$?
Yes. $\displaystyle a_i=x_i$ (not $\displaystyle x_1+x_2+...$) and $\displaystyle b_i=1$.

9. Thanks a lot for the help!

10. for question one , i attempt to prove this by Mathematical Induction .

When $\displaystyle n=1$

1^2 = $\displaystyle \frac{1}{x_1} \cdot ( x_{1} )$

Let $\displaystyle P = x_1 + x_2 + ... + x_k , Q = \frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_{k}}$

We assume $\displaystyle PQ \geq k^2$

When $\displaystyle n = k+1$

$\displaystyle ( x_1 + x_2 + ... + x_k + x_{k+1} ) \cdot ( \frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_{k}} + \frac{ 1}{x_{k+1}} )$ $\displaystyle = (P + x_{k+1} ) (Q + \frac{1}{x_{k+1}}) = $$\displaystyle PQ + 1 + \frac{P}{x_{k+1}} + x_{k+1} Q \geq k^2 + 1 + 2\sqrt{PQ}$$\displaystyle \geq k^2 + 2k + 1 = (k+1)^2$