Kantorovich Inequality: An inequality for real numbers I recently saw an inequality for real numbers known as Kantorovich inequality for real numbers. This says:
Suppose $x_1< x_2< \dots < x_n$ are positive real numbers and let let $\lambda_1, \lambda_2, \dots \lambda_n \geq 0$ with $\sum\limits_{j=1}^n \lambda_i=1$. Then
$$\left(\sum\limits_{j=1}^n \lambda_j x_j \right)\left(\sum\limits_{j=1}^n \lambda_j x_j^{-1} \right)\leq \dfrac{(x_1+x_n)^2}{4x_1x_n}.$$
My question is can we replace the strict inequality for $x_i$ by $``\leq"$? It seems so but I cannot prove it. Can anyone help me on this? By proving or by supporting documents?
Thanks in advance.
 A: Just one remark :
We have with the OP's constraints :
$$\left(\sum\limits_{j=1}^n \lambda_j x_j \right)\left(\sum\limits_{j=1}^n \lambda_j x_j^{-1} \right)\leq \left(\sum\limits_{j=1}^n \frac{1}{n} x_j \right)\left(\sum\limits_{j=1}^n \frac{1}{n} x_j^{-1} \right)\leq \dfrac{(x_1+x_n)^2}{4x_1x_n}.$$
With $\lambda_n\geq\cdots\geq\lambda_1$
Can you end now ?
A sketch for the LHS :
We start by prove the case $n=2$
The case $n=2$ have the form :
$$(\lambda_1a+(1-\lambda_1)b)(\lambda_2a+(1-\lambda_2)b)$$
Where $a+b=1$ .Remains to take the logarithm and use Jensen's inequality .
Now the case $n=3$
We have with $x_1\geq x_2\geq x_3>0$ and $\lambda_1\geq\lambda_2\geq\lambda_3>0$ and $\lambda_1+\lambda_2+\lambda_3=1$
$$\frac{4}{9}\frac{(\lambda_1x_1+(\lambda_2+\lambda_3)(x_2+x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2+\lambda_3}{x_2+x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})}\geq \frac{(\lambda_1x_1+\lambda_2x_2+\lambda_3x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2}{x_2}+\frac{\lambda_3}{x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})} $$
And :
$$\frac{(\lambda_1x_1+\frac{(1-\lambda_1)}{2}(x_2+x_3))(\frac{\lambda_1}{x_1}+\frac{(1-\lambda_1)}{2}\left(\frac{1}{x_2}+\frac{1}{x_3}\right))}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})}\geq \frac{(\lambda_1x_1+\lambda_2x_2+\lambda_3x_3))(\frac{\lambda_1}{x_1}+\frac{\lambda_2}{x_2}+\frac{\lambda_3}{x_3})}{(x_1+x_2+x_3)(\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3})} $$
This method works in the general case .
