Question

Saluut! Cum demostrez asta ? :
[tex](a1 + a2+ ...+an) ^{2} \leq n*(a1) ^{2} + n*(a2 )^{2} + n*(a3) ^{2} +...+ n*(an) ^{2}
[/tex]
Specific ca aceia sunt indici !

Answer 1

Din exemplul pentru 3 termani, se observa pentru cazul general fara inductie completa. $(a_{1}+ a_{2}+ a_{3})^2 \leq 3 a_{1}^2+3 a_{2}^2+ 3a_{3}^2$, demonstram folosind echivalenta, presupunem adevarata relatia, ea este echivalenta cu :$a_{1}^2 + a_{2}^2 + a_{3}^2+2( a_{1} a_{2}+ a_{1} a_{3}+ a_{2} a_{3}) \leq 3 a_{1}^2+ 3a_{2}^2+ 3 a_{3}^2$, trecem totul in dreapta si facem reducerile, ⇔ [tex]0 \leq 2 a_{1}^2+2a _{2}^2+ 2a_{3}^2-2 a_{1} a_{2}-2 a_{1} a_{2}-2 a_{2} a_{3} [/tex], echivalent cu: 0$\leq ( a_{1}- a_{2})^2 +( a_{1}- a_{3})^2 +( a_{2}- a_{3})^2$, evident adevarata ca suma de patrate, deci este ≥0, ceea ce inseamna ca presupunerea este adevarata. Procedand analog cu relatia data se obtine: $0 \leq (a_{1}- a_{2})^2+ (a_{1}- a_{3})^2+...+( a_{1}- a_{n})^2+ (a_{2}- a_{3})^2+...+$$( a_{2}- a_{n})^2+...+ (a_{n-1}- a_{n})^2,adevarata.$

Answer 2

$Nu~este~nevoie,~iata: \\ \\ Imparti~relatia~la~n^2~si~obtii: \\ \\ \frac{(a_1+a_2+...+a_n)^2}{n^2} \leq \frac{a_1^2+a_2^2+...+a_n^2}{n} . \\ \\ Termenii~fiind~ \geq 0,~ultima~relatie~este~echivalenta~cu: \\ \\ \frac{a_1+a_2+...+a_n}{n} \leq \sqrt{ \frac{a_1^2+a_2^2+...+a_n^2}{n} },~care~este~adevarata~conform \\ \\ inegalitatii~mediilor~(membrul~stang~este~media~aritmetica, \\ \\ iar~membrul~drept~este~media~patratica).$

$Sau,~in~fine,~inegalitatea~ta~este~de~fapt~un~caz \\ \\ particular~al~inegalitatii~Cauchy-Buniakovski-Schwartz. \\ \\ (a_1^2+a_2^2+...+a_n^2)(1^2+1^2+...+1^2) \geq (a_1 \cdot 1+a_2 \cdot 1 +...+ \\ \\ +a_n \cdot 1).$