This problem starts by giving us the result of the Cauchy-Schwarz Inequality. Now it's a matter of going from the two vector's inner products with themselves back to the original vectors. If $\langle v,v \rangle=n.$ Then $v$ must equal $\langle1,1,...1_n\rangle$ since its inner product equals its number of elements. Now if $\langle w,w \rangle=(a_1^2+a_2^2 ...+a_n^2)$, then $w$ must equal $\langle a_1,a_2,...a_n \rangle$ since its inner product is also the vector dotted with itself or the sum of its elements squared. Now the right side of the must be the inner product $|\langle v,w \rangle|^2$ which is $(\langle1,1,...1_n\rangle \cdot \langle a_1,a_2,a_2\rangle)^2$. Computing this inner product and squaring it gives the quantity $(a_1 + a_2... + a_n)^2$ which is the same as the left side of the original inequality. Since the Cauchy-Schwarz is defined for all real numbers this holds for all real values of $a$.
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This problem starts by giving us the result of the Cauchy-Schwarz Inequality. Now it's a matter of going from the two vector's inner products with themselves back to the original vectors. If $\langle v,v \rangle=n.$ Then $v$ must equal $\langle1,1,...1_n\rangle$ since its inner product equals its number of elements. Now if $\langle w,w \rangle=(a_1^2+a_2^2 ...+a_n^2)$, then $w$ must equal $\langle a_1,a_2,...a_n \rangle$ since its inner product is also the vector dotted with itself or the sum of its elements squared. Now the right side of the must be the inner product $|\langle v,w \rangle|^2$ which is $(\langle1,1,...1_n\rangle \cdot \langle a_1,a_2,a_2\rangle)^2$. Computing this inner product and squaring it gives the quantity $(a_1 + a_2... + a_n)^2$ which is the same as the left side of the original inequality. Since the Cauchy-Schwarz is defined for all real numbers this holds for all real values of $a$.