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

Michael Hardy
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Jamesbond
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    Welcome to MSE! Please let us know what you have tried so far and where you are stuck. – Tom Nov 01 '14 at 00:31
  • 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 01 '14 at 01:15

2 Answers2

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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 $a = x_1^2+...+x_n^2$.

DeepSea
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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*}

user2345215
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