I am reading Mathematical Tapas, here:

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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.

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

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:

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


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