Voi folosi inegalitatea mediilor:
[tex]m_p \geq m_a[/tex]
[tex]\sqrt{\dfrac{a^2+b^2+c^2}{3}} \geq \dfrac{a+b+c}{3} \Bigg|^2 \Rightarrow \dfrac{a^2+b^2+c^2}{3} \geq \dfrac{(a+b+c)^2}{9} \Rightarrow[/tex]
[tex]\Rightarrow 3(a^2+b^2+c^2) \geq (a+b+c)^2[/tex]
Astfel:
[tex]3(x_1^2+x_2^2+x_3^2) \geq (x_1+x_2+x_3)^2[/tex]
[tex]3(x_1^2+x_2^2+x_4^2) \geq (x_1+x_2+x_4)^2[/tex]
[tex]3(x_1^2+x_2^2+x_5^2) \geq (x_1+x_2+x_5)^2[/tex]
[tex]\vdots[/tex]
[tex]3(x_{n-2}^2+x_{n-1}^2+x_{n}^2) \geq (x_{n-2}+x_{n-1}+x_{n})^2[/tex]
Observăm că totalitatea x-șilor din suma din membrul drept a tuturor inegalităților reprezintă mulțimea formată din [tex]C_{n}^{3}[/tex] pentru {x₁; x₂; x₃; ...; xₙ}.
Dar x₁, x₂, ..., xₙ apar individual doar de [tex]C_{n-1}^{3-1}[/tex] ori în mulțimea combinărilor.
Adun inegalitățile:
[tex]\Rightarrow 3(x_1^2+x_2^2+x_3^2+x_1^2+x_2^2+x_4^2+...)\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2[/tex]
[tex]\Rightarrow 3(C_{n-1}^{3-1}x_1^2+C_{n-1}^{3-1}x_2^2+...+C_{n-1}^{3-1}x_n^2)\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2[/tex]
[tex]\Rightarrow 3C_{n-1}^{2}(x_1^2+x_2^2+...+x_n^2)\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2[/tex]
[tex]\Rightarrow 3\cdot \dfrac{(n-1)!}{2!(n-1-2)!}\cdot \sum\limits_{k=1}^nx_k^2\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2[/tex]
[tex]\Rightarrow 3\cdot \dfrac{(n-3)!(n-2)(n-1)}{2!(n-3)!}\cdot \sum\limits_{k=1}^nx_k^2\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2[/tex]
[tex]\Downarrow[/tex]
[tex]\boxed{\dfrac{3(n-1)(n-2)}{2}\cdot \sum\limits_{k=1}^nx_k^2\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2}[/tex]
