Question

problema 34 va rog!! multumesc✨

Answer 1

Voi folosi inegalitatea mediilor:

$m_p \geq m_a$

$\sqrt{\dfrac{a^2+b^2+c^2}{3}} \geq \dfrac{a+b+c}{3} \Bigg|^2 \Rightarrow \dfrac{a^2+b^2+c^2}{3} \geq \dfrac{(a+b+c)^2}{9} \Rightarrow$

$\Rightarrow 3(a^2+b^2+c^2) \geq (a+b+c)^2$

Astfel:

$3(x_1^2+x_2^2+x_3^2) \geq (x_1+x_2+x_3)^2$

$3(x_1^2+x_2^2+x_4^2) \geq (x_1+x_2+x_4)^2$

$3(x_1^2+x_2^2+x_5^2) \geq (x_1+x_2+x_5)^2$

$\vdots$

$3(x_{n-2}^2+x_{n-1}^2+x_{n}^2) \geq (x_{n-2}+x_{n-1}+x_{n})^2$

Observăm că totalitatea x-șilor din suma din membrul drept a tuturor inegalităților reprezintă mulțimea formată din $C_{n}^{3}$ pentru {x₁; x₂; x₃; ...; xₙ}.

Dar x₁, x₂, ..., xₙ apar individual doar de $C_{n-1}^{3-1}$ ori în mulțimea combinărilor.

Adun inegalitățile:

$\Rightarrow 3(x_1^2+x_2^2+x_3^2+x_1^2+x_2^2+x_4^2+...)\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2$

$\Rightarrow 3(C_{n-1}^{3-1}x_1^2+C_{n-1}^{3-1}x_2^2+...+C_{n-1}^{3-1}x_n^2)\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2$

$\Rightarrow 3C_{n-1}^{2}(x_1^2+x_2^2+...+x_n^2)\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2$

$\Rightarrow 3\cdot \dfrac{(n-1)!}{2!(n-1-2)!}\cdot \sum\limits_{k=1}^nx_k^2\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2$

$\Rightarrow 3\cdot \dfrac{(n-3)!(n-2)(n-1)}{2!(n-3)!}\cdot \sum\limits_{k=1}^nx_k^2\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2$

                                       $\Downarrow$

$\boxed{\dfrac{3(n-1)(n-2)}{2}\cdot \sum\limits_{k=1}^nx_k^2\geq \sum\limits_{1\leq k < i < j \leq n} (x_k+x_i+x_j)^2}$