I need some help with this please(difficult college homework)...
Prove that if $n \in \mathbb{N^*}$ and $t_1, t_2, ..., t_n \in \mathbb{ℝ^*_+}$ are such that $(\sum\limits_{k=1}^nt_k^2)(\sum\limits_{k=1}^n\frac{1}{t_k}) = 2(\sum\limits_{k=1}^nt_k)$ and$(\sum\limits_{k=1}^nt_k)(\sum\limits_{k=1}^n\frac{1}{t_k}) = \frac{3}{2}n^2$ then $\sum\limits_{k=1}^n\frac{t_k^3}{\sum\limits_{j=1,j\ne k}^nt_j^2} \ge \frac{n}{2}H(t_1,t_2,...,t_n)$, where $H(t_1, t_2, ...t_n)$ is the harmonic mean of $t_1, t_2,...,t_n$
I have spent almost a day on this problem. I first expanded right side of the inequality using the harmonic mean formula and identities I have, reaching to smth like this: $\frac{1}{3}\sum\limits_{k=1}^nt_k$. Then I pretended $\sum\limits_{k=1}^n\frac{t_k^3}{\sum\limits_{j=1,j\ne k}^nt_j^2}$ can be rewritten as $\sum\limits_{k=1}^n\frac{t_k^3}{\sum\limits_{j=1}^nt_j^2-t_k^2}$ which I said it's $\ge$ than $\sum\limits_{k=1}^n\frac{t_k^3}{\sum\limits_{j=1}^nt_j^2}$.I feel i did it wrong here, however that was the best I could do. Then I separated this sum in two independent sums and tried to find some inequality related to the right side of the main inequality. I tried with cube mean $\ge$ rms $\ge$ am but still nothing. I sense a modified version of Holder inequality or other super stuff like that might help, but I'm running out of energy so I would appreciate a little help
(Btw,I am new to TeX and math.stackexchange)
Thanks
Using the conditions and Holder's inequality: $$3n^2=\left(\sum t_k^2\right) \left(\sum \frac1{t_k}\right)^2 \ge n^3 \implies n \le 3$$
Now $n=1$ gives $t_1 \cdot \frac1{t_1} = \frac32$ which is not possible.
For $n=2$, the conditions can be written as for $x, y > 0$ $$(x^2+y^2)\left(\frac1x+\frac1y \right)=2(x+y), \quad (x+y)\left(\frac1x+\frac1y \right)=6$$ $$\implies 3(x^2+y^2) = (x+y)^2 \implies x^2+y^2=xy $$ which is not possible.
We are left with $n=3$, where we may write the condition for $x, y, z > 0$ $$(x^2+y^2+z^2)\left(\frac1x+\frac1y+\frac1z \right)=2(x+y+z), \quad (x+y+z)\left(\frac1x+\frac1y +\frac1z\right)=\frac{27}2$$ $$\implies 27(x^2+y^2+z^2)=4(x+y+z)^2$$ But from the well known inequality $3(x^2+y^2+z^2) \ge (x+y+z)^2$, the above is not possible.
Hence there is no $n$ for which the conditions are satisfied, hence trivially the result is true.