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Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu.

Data publicarii: 22 Noiembrie, 2008

TEORIE

Inegalitati uzuale:

  • {a^2 + b^2} \geq{ 2ab},\forall{a,b}\in{\mathbb{R}};{a^2 + b^2} \geq{ 2ab},\forall{a,b}\in{\mathbb{R}};  

(egalitate daca si numai daca a = b).

  • {a^2 + b^2 + c^2}\geq{ ab + bc + ca},\forall{a,b,c}\in{\mathbb{R}};{a^2 + b^2 + c^2}\geq{ ab + bc + ca},\forall{a,b,c}\in{\mathbb{R}};

(egalitate daca si numai daca a = b = c).

  • |\frac{a}{b} + \frac{b}{a}|\geq2,\forall{a,b}\in{\mathbb{R}};|\frac{a}{b} + \frac{b}{a}|\geq2,\forall{a,b}\in{\mathbb{R}};

(egalitate daca si numai daca a = +b, sau a = -b).

  • |{x_1}+{x_2}+\cdots+{x_n}|\leq{|{x}_{1}|}+{|{x}_{2}|}+...+{|{x}_{n}|},\forall{x_k}\in{\mathbb{R}},\forall{k}\in{{\mathbb{N}}^*};|{x_1}+{x_2}+\cdots+{x_n}|\leq{|{x}_{1}|}+{|{x}_{2}|}+...+{|{x}_{n}|},\forall{x_k}\in{\mathbb{R}},\forall{k}\in{{\mathbb{N}}^*};

(egalitate pentru n = 1, sau xi·xЄ[0,+oo) pentru orice i,jЄ{1,2, ...,n}).

  • {|a|}\leq{c}\Leftrightarrow -{c}\leq{a} \leq{c},\forall{a}\in{\mathbb{R}},\forall{c} > 0.{|a|}\leq{c}\Leftrightarrow -{c}\leq{a} \leq{c},\forall{a}\in{\mathbb{R}},\forall{c} > 0.
  • {|a|}\geq{c}\Leftrightarrow {a}\in{(-\infty,-c]\cup[c,+\infty)},\forall{a}\in{\mathbb{R}},\forall{c} > 0.{|a|}\geq{c}\Leftrightarrow {a}\in{(-\infty,-c]\cup[c,+\infty)},\forall{a}\in{\mathbb{R}},\forall{c} > 0.
  • {|acosx+bsinx|}\le{\sqrt{a^2+b^2}},\;\forall{a,b,x}\in{\mathbb{R}}.{|acosx+bsinx|}\le{\sqrt{a^2+b^2}},\;\forall{a,b,x}\in{\mathbb{R}}.
  • |x| ≤ |tgx|, oricare ar fi x Є (-π/2;π/2).
  • {2}^{n}>{n},\forall{n}\in{\mathbb{N}}.{2}^{n}>{n},\forall{n}\in{\mathbb{N}}.
  • {{\frac{2}{3}}\cdot{\frac{4}{5}}\cdot{\frac{6}{7}}\cdots{\frac{2n}{2n+1}}}<{\frac{1}{\sqrt{n+1}}},\;\forall{n}\in{\mathbb{N^*}}.{{\frac{2}{3}}\cdot{\frac{4}{5}}\cdot{\frac{6}{7}}\cdots{\frac{2n}{2n+1}}}<{\frac{1}{\sqrt{n+1}}},\;\forall{n}\in{\mathbb{N^*}}.
  • {{\frac{1}{2}}\cdot{\frac{3}{4}}\cdot{\frac{5}{6}}\cdots{\frac{2n-1}{2n}}}<{\frac{1}{\sqrt{2n+1}}},\;\forall{n}\in{\mathbb{N^*}}.{{\frac{1}{2}}\cdot{\frac{3}{4}}\cdot{\frac{5}{6}}\cdots{\frac{2n-1}{2n}}}<{\frac{1}{\sqrt{2n+1}}},\;\forall{n}\in{\mathbb{N^*}}.
  • {1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}>{ln(n+1)},\forall{n}\in{\mathbb{N^*}}.{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}>{ln(n+1)},\forall{n}\in{\mathbb{N^*}}.
  • {e^x}\ge{x+1},\;\forall{x}\in{\mathbb{R}}.{e^x}\ge{x+1},\;\forall{x}\in{\mathbb{R}}.
  • {lnx}<{x}<{e^x},\;\forall{x}\ge{1}.{lnx}<{x}<{e^x},\;\forall{x}\ge{1}.
  • {log_ab+lob_ba}>{2};\;(a>1\;si\;b>1)\;sau\;({0}<{a}<{1},\;{0}<{b}<{1});\;a\not=b.{log_ab+lob_ba}>{2};\;(a>1\;si\;b>1)\;sau\;({0}<{a}<{1},\;{0}<{b}<{1});\;a\not=b.

Inegalitatile mediilor(armonicageometrica≤aritmeticapatratica):

{min(a,b)}\leq{\frac{2ab}{a+b}}\leq{\sqrt{ab}}\leq{\frac{a+b}{2}}\leq{\sqrt{\frac{a^2+b^2}{2}}}\leq{ max(a,b)},\forall{a,b}\in{(0,\infty)};{min(a,b)}\leq{\frac{2ab}{a+b}}\leq{\sqrt{ab}}\leq{\frac{a+b}{2}}\leq{\sqrt{\frac{a^2+b^2}{2}}}\leq{ max(a,b)},\forall{a,b}\in{(0,\infty)};  

(egalitate daca si numai daca a = b). 

Generalizare:

{min({x_1},\cdots,{x_n})}\leq{\frac{n}{\frac{1}{{x_1}}+\cdots+\frac{1}{{x_n}}}\leq{\sqrt[n]{{x_1}\cdots{x_n}}}\leq{\frac{{x_1}+\cdots+{x_n}}{n}}}\leq{min({x_1},\cdots,{x_n})}\leq{\frac{n}{\frac{1}{{x_1}}+\cdots+\frac{1}{{x_n}}}\leq{\sqrt[n]{{x_1}\cdots{x_n}}}\leq{\frac{{x_1}+\cdots+{x_n}}{n}}}\leq {\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}}}\leq{\sqrt{\frac{x_1^2+\cdots+x_n^2}{n}}}\leq \leq{max({x_1},\cdots,{x_n})},\forall{x_k}>{0},\forall{n}>{1};\leq{max({x_1},\cdots,{x_n})},\forall{x_k}>{0},\forall{n}>{1};

(egalitate daca si numai daca x1 = x2  = ... = xn).

Inegalitatea Cauchy-Schwarz:

{(\sum_{k=1}^{k=n}{{a}_{k}}{{b}_{k}})}^{2}\leq{(\sum_{k=1}^{k=n}{{{a}_{k}}^{2}})}{(\sum_{k=1}^{k=n}{{{b}_{k}}^{2}})},\forall{{a_k},{b_k}}\in{\mathbb{R}},\forall{n}\in{{\mathbb{N}}^*};{(\sum_{k=1}^{k=n}{{a}_{k}}{{b}_{k}})}^{2}\leq{(\sum_{k=1}^{k=n}{{{a}_{k}}^{2}})}{(\sum_{k=1}^{k=n}{{{b}_{k}}^{2}})},\forall{{a_k},{b_k}}\in{\mathbb{R}},\forall{n}\in{{\mathbb{N}}^*};

(egalitate daca si numai daca exista numarul real r, astfel incat ak = r·bk, oricare ar fi

kЄ{1,2, ... ,n}).

Inegalitatea lui Minkovschi:

\sqrt{\sum_{k=1}^{k=n}{{({a}_{k}+{b}_{k})}^{2}}}\leq\sqrt{\sum_{k=1}^{k=n}{{{a}_{k}}^{2}}}+\sqrt{\sum_{k=1}^{k=n}{{{b}_{k}}^{2}}},\sqrt{\sum_{k=1}^{k=n}{{({a}_{k}+{b}_{k})}^{2}}}\leq\sqrt{\sum_{k=1}^{k=n}{{{a}_{k}}^{2}}}+\sqrt{\sum_{k=1}^{k=n}{{{b}_{k}}^{2}}}, \forall{{a_k},{b_k}}\in{\mathbb{R}};\forall{{a_k},{b_k}}\in{\mathbb{R}};

(egalitate daca si numai daca exista numarul real r, astfel incat ak = r·bk, oricare ar fi

kЄ{1,2, ... ,n}).

Inegalitatea lui Cebasev:

I) Daca (ak), (bk) sunt doua siruri monotone, avand acelasi sens de monotonie

(ambele crescatoare sau ambele descrescatoare), atunci:

{\frac{\sum_{k=1}^{k=n}{{a}_{k}}}{n}}\cdot{\frac{\sum_{k=1}^{k=n}{{b}_{k}}}{n}}\leq{\frac{\sum_{k=1}^{k=n}{{{a}_{k}}\cdot{{b}_{k}}}}{n}},\forall{n}\in{\mathbb{N}}^{*},\forall{{a}_{k},{b}_{k}}\in{\mathbb{R}}.{\frac{\sum_{k=1}^{k=n}{{a}_{k}}}{n}}\cdot{\frac{\sum_{k=1}^{k=n}{{b}_{k}}}{n}}\leq{\frac{\sum_{k=1}^{k=n}{{{a}_{k}}\cdot{{b}_{k}}}}{n}},\forall{n}\in{\mathbb{N}}^{*},\forall{{a}_{k},{b}_{k}}\in{\mathbb{R}}.

II) Daca (ak), (bk) sunt doua siruri monotone, avand sensuri diferite de monotonie

(unul crescator, celalalt descrescator), atunci:

{\frac{\sum_{k=1}^{k=n}{{a}_{k}}}{n}}\cdot{\frac{\sum_{k=1}^{k=n}{{b}_{k}}}{n}}\geq{\frac{\sum_{k=1}^{k=n}{{{a}_{k}}\cdot{{b}_{k}}}}{n}},\forall{n}\in{\mathbb{N}}^{*},\forall{{a}_{k},{b}_{k}}\in{\mathbb{R}}.{\frac{\sum_{k=1}^{k=n}{{a}_{k}}}{n}}\cdot{\frac{\sum_{k=1}^{k=n}{{b}_{k}}}{n}}\geq{\frac{\sum_{k=1}^{k=n}{{{a}_{k}}\cdot{{b}_{k}}}}{n}},\forall{n}\in{\mathbb{N}}^{*},\forall{{a}_{k},{b}_{k}}\in{\mathbb{R}}.

Inegalitatea lui Bernoulli:

{(1+x)}^{\alpha}\geq{1+{\alpha}\cdot{x}},{(1+x)}^{\alpha}\geq{1+{\alpha}\cdot{x}},

pentru\;{x}>{-1},\;{\alpha}>{1},\;sau\;{\alpha}<{0}.pentru\;{x}>{-1},\;{\alpha}>{1},\;sau\;{\alpha}<{0}.

\quad{(1+x)}^{\alpha}\leq{1+{\alpha}\cdot{x}},\quad{(1+x)}^{\alpha}\leq{1+{\alpha}\cdot{x}},

pentru\;{x}>{-1},\;{0}<{\alpha}<{1},pentru\;{x}>{-1},\;{0}<{\alpha}<{1},

(cu egalitate in ambele inegalitati daca si numai daca x = 0).

Inegalitatea lui Jensen:

I) Daca functia f:I - > R este convexa, adica 

\forall{{x_1},{x_2}}\in{I},\forall{t}\in{[0,1]}:\forall{{x_1},{x_2}}\in{I},\forall{t}\in{[0,1]}:

{f({(1-t)}{{x_1}}+{t}\cdot{{x_2}})}\leq (1-t)f({x_1})+tf({x_2}),{f({(1-t)}{{x_1}}+{t}\cdot{{x_2}})}\leq (1-t)f({x_1})+tf({x_2}),

atunci: 

{f(\sum_{k=1}^{k=n}{{q_k}{x_k}})}\leq{\sum_{k=1}^{k=n}{{q_k}{f(x_k)}}},{f(\sum_{k=1}^{k=n}{{q_k}{x_k}})}\leq{\sum_{k=1}^{k=n}{{q_k}{f(x_k)}}},

unde 

qk > 0; k = 1,2,...,n; q1 + q2 + ... + qn = 1 si 

\sum_{k=1}^{k=n}{{q_k}}\cdot{{x_k}}\in{I},\sum_{k=1}^{k=n}{{q_k}}\cdot{{x_k}}\in{I},  

pentru xЄI. 

Caz particular: 

Daca: 

q_i=\frac{1}{n},\forall{i}\in{\overline{1,n}}q_i=\frac{1}{n},\forall{i}\in{\overline{1,n}}

si functia f:I - > R este convexa, atunci:

{f(\frac{{x_1}+{x_2}+...+{x_n}}{n})}\leq\frac{f({x_1})+f({x_2})+...+f({x_n})}{n},\forall{{x_k}}\in{I},{k=1,2,...,n}.{f(\frac{{x_1}+{x_2}+...+{x_n}}{n})}\leq\frac{f({x_1})+f({x_2})+...+f({x_n})}{n},\forall{{x_k}}\in{I},{k=1,2,...,n}.

II) Daca functia f:I - > R este concava, adica pentru orice x1,xЄI si orice tЄ[0,1] avem

{f({(1-t)}{{x_1}}+{t}\cdot{{x_2}})}\geq (1-t)f({x_1})+tf({x_2}),{f({(1-t)}{{x_1}}+{t}\cdot{{x_2}})}\geq (1-t)f({x_1})+tf({x_2}),

atunci: 

{f(\sum_{k=1}^{k=n}{{q_k}{x_k}})}\geq{\sum_{k=1}^{k=n}{{q_k}{f(x_k)}}},{f(\sum_{k=1}^{k=n}{{q_k}{x_k}})}\geq{\sum_{k=1}^{k=n}{{q_k}{f(x_k)}}},

unde qk > 0; k = 1, 2, ... , n, q1+q2+...+q= 1 si 

\sum_{k=1}^{k=n}{{q_k}}\cdot{{x_k}}\in{I},\sum_{k=1}^{k=n}{{q_k}}\cdot{{x_k}}\in{I},  

pentru xЄI.

Caz particular:

Daca: 

q_i=\frac{1}{n},\forall{i}\in{\overline{1,n}}q_i=\frac{1}{n},\forall{i}\in{\overline{1,n}}

si functia f:I - > R este concava, atunci:

{f(\frac{{x_1}+{x_2}+...+{x_n}}{n})}\geq\frac{f({x_1})+f({x_2})+...+f({x_n})}{n},\forall{{x_k}}\in{I},{k=1,2,...,n}.{f(\frac{{x_1}+{x_2}+...+{x_n}}{n})}\geq\frac{f({x_1})+f({x_2})+...+f({x_n})}{n},\forall{{x_k}}\in{I},{k=1,2,...,n}.

Postat în: INEGALITATI-liceu

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Răspunsuri şi comentarii

Multumesc

Costel, 19.10.2015 17:24

multumesc chiar ma ajutat respectele mele

Răspuns: Cu multa placere! Ma bucur!

 

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