Let $n\ge 100$ be an integer. Let $x_1,x_2,\dots ,x_n$ be positive real numbers such that $x_1+x_2+\cdots +x_n=1$. Prove that:

Step 1: Apply Cauchy's Engel Form (also known as Titu's Lemma), which states that for positive real numbers $a_i$ and $b_i$:

Step 2: Let $a_i=x_i$ and $b_i=x_i+x_{i+1}$ (with index wrapping $x_{n+1}=x_1$). Applying the lemma:

Step 3: Simplify the terms in the numerator and denominator:

• Numerator: $(\sum x_i{)}^2=1^2=1$.

• Denominator: ${\sum }_{i=1}^n(x_i+x_{i+1})=2{\sum }_{i=1}^n{x}_i=2(1)=2$.

Step 4: Substitute these simplifications back into the inequality:

This completes the proof. Equality holds if and only if $x_1=x_2=\cdots =x_n=1/n$.