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»INEGALITATI

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 · xj € [0, + oo) pentru orice i, j € {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}}.
${2}^{n}>{n},\forall{n}\in{\mathbb{N}}.$ {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:

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

(egalitate daca si numai daca a = b).

Generalizare:

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

(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

k din multimea {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

k din multimea {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, definita pe intervalul I si cu valori in 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, pentru orice xk din I, atunci:

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

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, definita pe intervalul I si cu valori in 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, definita pe intervalul I si cu valori in R, este concava, adica

pentru orice x1, x2 din I si orice t din [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 + ... + qn = 1, pentru orice xk din I, atunci:

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

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, definita pe intervalul I si cu valori in R, este concava pe intervalul, 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}.

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