Real Number recursion

Question

Let $T_{n}$ be a sequence of real numbers such that $T_{n+1}\ge T_{n}^{2}+\frac{1}{5}$ Prove that $\sqrt{T_{n+5}}\ge T_{n}$. I rearranged the first expression in terms of $T_{n}$, but seemingly to no avail. Are there any other pointers for solving this problem?

Answer 1

For all real $x$ we have $$ x^2-x = \Bigl(x-{\small{\frac{1}{2}}}\Bigr)^2 - {\small{\frac{1}{4}}} \ge -{\small{\frac{1}{4}}} $$ so then we get \begin{align*} T_{n+5} &= T_{n+1} + \sum_{k=1}^4 (T_{n+k+1}-T_{n+k}) \\[4pt] &\ge T_{n+1} + \sum_{k=1}^4 \left(\Bigl(T_{n+k}^2+{\small{\frac{1}{5}}}\Bigr)-T_{n+k}\right) \\[4pt] &= T_{n+1} + \sum_{k=1}^4 \left(\Bigl(T_{n+k}^2-T_{n+k}\Bigr)+{\small{\frac{1}{5}}}\right) \\[4pt] &\ge T_{n+1} + \sum_{k=1}^4 \Bigl(-{\small{\frac{1}{4}}}+{\small{\frac{1}{5}}}\Bigr) \\[4pt] &= T_{n+1} + \sum_{k=1}^4 \Bigl(-{\small{\frac{1}{20}}}\Bigr) \\[4pt] &= T_{n+1} - {\small{\frac{1}{5}}} \\[4pt] &\ge \Bigl(T_n^2+{\small{\frac{1}{5}}}\Bigr) - {\small{\frac{1}{5}}} \\[4pt] &= T_n^2 \\[4pt] \end{align*} hence $\sqrt{T_{n+5}}\ge T_n$.