Prove using squared number property

Question

$$ If \sum_{i=1}^{10} x_i=10 $$ Prove that $$ \sum_{i=1}^{10} x_i^2\ge 10 $$

Answer 1

By C-S $$\sum_{i=1}^{10}1^2\sum_{i=1}^{10}x_i^2\geq\left(\sum_{i=1}^{10}x_i\right)^2=100.$$ Id est, $$\sum_{i=1}^{10}x_i^2\geq10$$ and we are done!

Also, we can use the Tangent Line method: $$\sum_{i=1}^{10}x_i^2-10=\sum_{i=1}^{10}(x_i^2-1)=\sum_{i=1}^{10}(x_i^2-1-2(x_i-1))=$$ $$=\sum_{i=1}^{10}(x_i^2-2x_i+1)=\sum_{i=1}^{10}(x_i-1)^2\geq0.$$

Answer 2

Use the AM-GM.

For $n=2$ : $$x_1+x_2=10 \Rightarrow x_1^2+x_2^2+2x_1x_2=100\le 2(x_1^2+x_2^2) \Rightarrow x_1^2+x_2^2\ge \frac{100}{2}.$$ Similarly, for any $n$: $$x_1+x_2+\cdots +x_n=10 \Rightarrow \sum_{k=1}^nx_k^2+2\sum_{1\le i<j\le n}x_ix_j=100\le n\sum_{k=1}^nx_k^2 \Rightarrow \\\sum_{k=1}^nx_k^2\ge \frac{100}{n}.$$

Answer 3

By RMS-AM, $$\begin{align}\left(\frac1{10}\sum_{i=1}^{10}x_i^2\right)^{\frac12}&\geq\frac1{10}\sum_{i=1}^{10}x_i=1\\ \sum_{i=1}^{10}x_i^2&\geq10 \end{align}$$

Answer 4

Hint: for each $i$ we have $$x_i^2\geq 2x_i-1$$ and thus a conclusion.