# An Inequality problem 123 [closed]

Let $x_1,x_2,\ldots,x_n\in \left [ 0,1 \right ]$, prove that $(1+x_{1}+x_2+\cdots+x_n)^2\geq 4(x_1^2+x_2^2+\cdots+x_n^2)$

## closed as off-topic by Carl Mummert, gnometorule, Micah, John, Milo BrandtNov 5 '14 at 0:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, gnometorule, Micah, John, Milo Brandt
If this question can be reworded to fit the rules in the help center, please edit the question.

• Welcome to MSE! Please let us know what you have tried so far and where you are stuck. – Tom Nov 1 '14 at 0:31
• Use that $(1-x)^2\ge0$ hence $1+x^2\ge2x$. So, $(1+x)^2=1+2x+x^2\ge 2x+2x=4x\ge4x^2$, which proves the inequality when $n=1$. Then perhaps induction might work (I did not try). – Mirko Nov 1 '14 at 1:15

Since $x_i \in [0,1]$, we have: $1\geq x_i\geq x_i^2\geq 0$, thus:
$x_1+x_2+...+x_n \geq x_1^2+x_2^2+...+x_n^2\geq 0 \longrightarrow (1+x_1+...+x_n)^2\geq (1+x_1^2+...+x_n^2)^2 \geq 4(x_1^2+...+x_n^2)$ by the well-known inequality: $(1+a)^2 - 4a = (1-a)^2 \geq 0$ is true for $a = x_1^2+...+x_n^2$.
Suppose $1\ge x_1\ge x_2\ge \ldots \ge x_n\ge 0$, then \begin{align*}(1+x_1+x_2+\ldots+x_n)^2&=1+\sum_i x_i^2+2\sum_i x_i+\sum_{i\ne j} x_ix_j\\&\ge x_1^2+\sum_i x_i^2+2\sum_i x_i^2+\sum_{i\ne n}x_i x_{i+1}\\&\ge3\sum_i x_i^2+x_1^2+\sum_{i\ne n} x_{i+1}^2=4\sum_i x_i^2\end{align*}