# Inequality : $(t_1+t_n)^2\Bigg(\sum_{k=1}^n t_k\Bigg)^2 \geq 4nt_1t_n \Bigg(\sum_{k=1}^n t_k^2\Bigg)$ for $t_1 \leq t_2 \leq \ldots \leq t_n$

I conjecture (and have shown for $n\leq 3$) the following : if $t_1 \leq t_2 \leq \ldots \leq t_n$ are nonnegative numbers, then

$$(t_1+t_n)^2 \Bigg(\sum_{k=1}^n t_k\Bigg)^2 \geq 4nt_1t_n \Bigg(\sum_{k=1}^n t_k^2\Bigg)$$

Any ideas on how to prove this conjecture or find a counterexample ?

My progress so far : for $n=3$, denote by $D$ the difference between the LHS and the RHS. Then, $(t_3-t_1)^2D$ can be written as

$$\begin{array}{l} \Bigg(2t_1(t_2^2-t_2t_3+t_3^2)-t_2(t_1^2+t_3^2)-(t_3-t_1)^3\Bigg)^2\\ +4t_1(t_3-t_2)(t_2-t_1)\Bigg((t_3-t_2)(t_3^2-t_1t_2)+(t_3-t_1)(t_3^2-t_1^2)\Bigg) \end{array}$$

• Let $n, t_1, t_n, \sum_{k=1}^n t_k$ fixed. Then we can maximize $\sum_{k=1}^n t_k^2$ by setting all but one $t_i$s equal to $t_1$ and $t_n$. – didgogns May 14 '17 at 16:15

From Pólya-Szegö’s inequality, we have for $0 < m_1 \leqslant u_k \leqslant M_1$ and $0 < m_2 \leqslant v_k \leqslant M_2$, $$\left(\sum u_k^2 \right) \left( \sum v_k^2 \right) \leqslant \frac14 \left( \sqrt{\frac{M_1 M_2}{m_1m_2}} + \sqrt{\frac{m_1 m_2}{M_1 M_2}} \right)^2 \left( \sum u_k v_k\right)^2$$
Taking $u_k = t_k$ and $v_k = 1$, with a little simplification, we have your inequality.
• Do you use the $t_1\leq t_2\leq \ldots t_n$ hypothesis somewhere ? – Ewan Delanoy May 14 '17 at 17:58
• @EwanDelanoy $t_k \in [t_1, t_n]$ is used, though not the complete ordering. Let me see if I can dig up a good reference on Pólya-Szegö’s inequality - should be fairly well known. – Macavity May 14 '17 at 18:00