# Thread: Interesting problem, any proofs?

1. ## Interesting problem, any proofs?

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

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

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

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

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

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

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

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

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:

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

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

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 $a_i$ and $b_i$ 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 $a_i$ and $b_i$ 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:

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

Now rearrange your problem a bit:

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

Do you see what to do now?

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

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

9. Thanks a lot for the help!

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

When $n=1$

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

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

We assume $PQ \geq k^2$

When $n = k+1$

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