1
$\begingroup$

I am reading Mathematical Tapas, here:

enter image description here

I have answered these two problems. The trouble is that I couldn't answer the second question in each of those. I have employed Cauchy-Schwarz inequality to answer them. I know that

$$|\langle \mathbf{u},\mathbf{v}\rangle| = \|\mathbf{u}\| \|\mathbf{v}\|$$

iff $\mathbf{u}$ is linearly dependent of $\mathbf{v}$. So, for example: In for the first one, I did $u_i=\sqrt{x_i}$ and $v_i=\frac{1}{\sqrt{x_i}}$, I think this means that we need to have:

$$(\sqrt{x_1},\dots,\sqrt{x_n})=\lambda\left(\frac{1}{\sqrt{x_1}} ,\dots , \frac{1}{\sqrt{x_n}}\right)$$

And hence:

$$\left(\sqrt{x_1}-\lambda\frac{1}{\sqrt{x_1}} ,\dots , \sqrt{x_n}-\lambda \frac{1}{\sqrt{x_n}}\right)=0$$

But this doesn't seems to be helpful.

$\endgroup$
1
  • $\begingroup$ For attaining $n^2$ you can take $x_k=\frac1n$ for every $k$ right? $\endgroup$
    – drhab
    Feb 1 '20 at 9:14
1
$\begingroup$

Cauchy-Schwarz has the equality case when the ratio of each corresponding term is constant. In your case that means:

$$\frac{\sqrt{x_k}}{\frac{1}{\sqrt{x_k}}} = \text{same constant},\ k = \overline{1,n}$$

This means $x_1=x_2=\ldots=x_n$. To find this constant value, just replace all the variables in the given restriction:

$$nx_1=1\Rightarrow x_1=\frac{1}{n}$$

So the equality case is $x_1=x_2=\ldots=x_n = \dfrac{1}{n}$. And this indeed reaches the required lower bound $n^2$. For the second inequality the same reasoning gives the equality case:

$$(x_1,x_2,\ldots,x_n) = \left(\frac{\mu}{a_1},\frac{\mu}{a_2},\ldots,\frac{\mu}{a_n}\right)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.