Given $a_{i}\in \left [ 0;1 \right ]$, $i=1,2,...,2019.$ Prove that: $$\left(\frac{\sum a_{i}+1}{2}\right)^{2}\geq \sum a_{i}^{2}.$$
I think it is a very fantastic inequality. From $a_{i}\in \left [ 0;1 \right ]$, I only get $a_i^2 \leq a_i$. I made an effort to use AM-GM but it is not effective. What is the mothod to solve this kind of problem? Give me the solution or any great ideas please, thank you in advance!
$$\sum_{i=1}^{2019}a_i^2\leq\sum_{i=1}^{2019}a_i$$ and we need to prove that $$\left(\frac{\sum\limits_{i=1}^{2019}a_i+1}{2}\right)^2\geq\sum_{i=1}^{2019}a_i,$$ which is true by AM-GM: $$\left(\frac{\sum\limits_{i=1}^{2019}a_i+1}{2}\right)^2\geq\left(\frac{2\sqrt{\sum\limits_{i=1}^{2019}a_i}}{2}\right)^2=\sum_{i=1}^{2019}a_i.$$ I used the following.
Let $\sum\limits_{i=1}^{2019}a_i=x$.
Thus, we need to prove that $$\left(\frac{x+1}{2}\right)^2\geq x$$ and by AM-GM: $$\left(\frac{x+1}{2}\right)^2\geq\left(\frac{2\sqrt{x\cdot1}}{2}\right)^2=x.$$