NIMCET 2022 — Mathematics PYQ

Explanation using Cauchy-Schwarz Inequality 
The relationship between the sum of squares and the square of the sum of real numbers can be established using the Cauchy-Schwarz inequality. 
Step 1: State the Cauchy-Schwarz Inequality 
For any real numbers $a_1,a_2,\dots ,a_n$ and $b_1,b_2,\dots ,b_n$, the Cauchy-Schwarz inequality states that $({\sum }_{i=1}^n{a}_i{b}_i{)}^2\le ({\sum }_{i=1}^n{a}_i^2)({\sum }_{i=1}^n{b}_i^2)$. 
Step 2: Apply the Inequality 
Let $b_i=1$ for all $i=1,2,\dots ,n$. Substituting this into the Cauchy-Schwarz inequality, it is obtained that $({\sum }_{i=1}^n{a}_i\cdot 1{)}^2\le ({\sum }_{i=1}^n{a}_i^2)({\sum }_{i=1}^n{1}^2)$. 
Step 3: Simplify the Expression 
The inequality can be simplified as $({\sum }_{i=1}^n{a}_i{)}^2\le ({\sum }_{i=1}^n{a}_i^2)(n)$. 
Step 4: Rearrange the Inequality 
Rearranging the terms, it is found that $n{\sum }_{i=1}^n{a}_i^2\ge ({\sum }_{i=1}^n{a}_i{)}^2$. 
Final Answer 
The correct option is (b) $n{\sum }_{i=1}^n{a}_i^2\ge ({\sum }_{i=1}^n{a}_i{)}^2$.