{"id":42993,"date":"2015-07-27T22:47:54","date_gmt":"2015-07-28T01:47:54","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=42993"},"modified":"2023-03-22T23:32:37","modified_gmt":"2023-03-23T02:32:37","slug":"problema-medias-pra-que-vos-quero","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-medias-pra-que-vos-quero\/","title":{"rendered":".Problem\u00e3o: M\u00e9dias, para que vos quero?"},"content":{"rendered":"

\n

\n

Problema
\n (Indicado a partir do 3\u00ba ano do E. M.)<\/span><\/p>\n

\nProve as afirma\u00e7\u00f5es abaixo.
\n(a) Para quaisquer [tex]x[\/tex] e [tex]y[\/tex] reais positivos temos [tex]\\left(\\dfrac{x+y}{2}\\right)^2 \\leq \\dfrac{x^2 + y^2}{2}[\/tex]; ocorrendo a igualdade se, e somente se, [tex]x=y[\/tex].<\/p>\n

(b) Se [tex]a[\/tex] e [tex]b[\/tex] s\u00e3o reais positivos tais que [tex]a + b =1[\/tex], ent\u00e3o [tex]\\left(a + \\dfrac{1}{a}\\right)^2 + \\left(b + \\dfrac{1}{b}\\right)^2 \\geq \\dfrac{25}{2}[\/tex]<\/p>\n

\n<\/div>\n

\n

\n

Solu\u00e7\u00e3o<\/p>\n

\nIndicaremos a m\u00e9dia aritm\u00e9tica e a m\u00e9dia geom\u00e9trica de dois n\u00fameros reais positivos [tex]a[\/tex] e [tex]b[\/tex] por [tex]A(a , b)[\/tex] e [tex]G(a , b)[\/tex] respectivamente.<\/p>\n

[tex](a)[\/tex] Pela desigualdade entre as m\u00e9dias aritm\u00e9tica e geom\u00e9trica, segue que:<\/p>\n

[tex]\\qquad G\\left(\\dfrac{x^2}{2}\\, , \\,\\dfrac{y^2}{2}\\right) \\leq A\\left(\\dfrac{x^2}{2}\\, , \\,\\dfrac{y^2}{2}\\right)[\/tex]<\/p>\n

[tex]\\qquad \\sqrt{\\dfrac{x^2}{2} \\cdot \\dfrac{y^2}{2}} \\leq \\dfrac{1}{2} \\cdot \\left( \\dfrac{x^2}{2} + \\dfrac{y^2}{2}\\right)[\/tex]<\/p>\n

[tex]\\qquad \\dfrac{xy}{2} \\leq \\dfrac{x^2 + y^2}{4}[\/tex]<\/p>\n

[tex]\\qquad \\dfrac{x^2 + y^2}{4} + \\dfrac{xy}{2} \\leq \\dfrac{x^2 + y^2}{4} + \\dfrac{x^2 + y^2}{4}[\/tex]<\/p>\n

[tex]\\qquad \\left( \\dfrac{x}{2}\\right)^2 + 2 \\cdot \\dfrac{x}{2} \\cdot \\dfrac{y}{2} + \\left( \\dfrac{y}{2}\\right)^2 \\leq \\dfrac{x^2 + y^2}{2}[\/tex]<\/p>\n

[tex]\\qquad \\left( \\dfrac{x+y}{2}\\right)^2 \\leq \\dfrac{x^2 + y^2}{2}[\/tex]<\/p>\n

Ocorrendo a igualdade se, e somente se, [tex]\\dfrac{x^2}{2} = \\dfrac{y^2}{2}[\/tex], ou seja, [tex]x = y[\/tex].<\/p>\n

\n

[tex](b)[\/tex] Usaremos o resultado anterior, tomando [tex]x = a + \\frac{1}{a}\\, [\/tex] e [tex]\\, y = b + \\frac{1}{b}[\/tex]. Assim:<\/p>\n

[tex]\\qquad \\dfrac{\\left(a + \\frac{1}{a}\\right)^2 + \\left(b + \\frac{1}{b}\\right)^2}{2} \\geq \\left(\\dfrac{a + \\frac{1}{a} + b + \\frac{1}{b}}{2}\\right)^2.\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(I)[\/tex]
\nMas [tex]a+b=1[\/tex], logo<\/p>\n

[tex]\\qquad \\left(\\dfrac{a + \\frac{1}{a} + b + \\frac{1}{b}}{2}\\right)^2= \\left(\\dfrac{1 + \\frac{1}{a} + \\frac{1}{b}}{2}\\right)^2[\/tex]<\/p>\n

[tex]\\qquad \\left(\\dfrac{a + \\frac{1}{a} + b + \\frac{1}{b}}{2}\\right)^2=\\dfrac{1}{4} \\cdot \\left(1 + \\dfrac{a + b}{ab}\\right)^2[\/tex]<\/p>\n

[tex]\\qquad \\left(\\dfrac{a + \\frac{1}{a} + b + \\frac{1}{b}}{2}\\right)^2 =\\dfrac{1}{4} \\cdot \\left(1 + \\dfrac{1}{ab}\\right)^2.\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(II)[\/tex]<\/p>\n

Por [tex](I)\\, [\/tex] e [tex]\\, (II)[\/tex], temos que
\n[tex]\\qquad \\dfrac{\\left(a + \\frac{1}{a}\\right)^2 + \\left(b + \\frac{1}{b}\\right)^2}{2} \\geq \\dfrac{1}{4} \\cdot \\left(1 + \\dfrac{1}{ab}\\right)^2.\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(III)[\/tex]<\/p>\n

\n

Pela desigualdade entre as m\u00e9dias geom\u00e9trica e aritm\u00e9tica, temos que<\/p>\n

[tex]\\qquad G(a,b) \\leq A(a,b) \\\\
\n\\qquad \\sqrt{ab} \\leq \\dfrac{a + b}{2}[\/tex]<\/p>\n

[tex]\\qquad \\qquad ab \\leq \\dfrac{1}{4}[\/tex]<\/p>\n

[tex]\\qquad \\qquad \\dfrac{1}{ab} \\geq 4[\/tex]<\/p>\n

[tex]\\qquad \\qquad 1 + \\dfrac{1}{ab} \\geq 5[\/tex]<\/p>\n

[tex]\\qquad \\qquad \\left(1 + \\dfrac{1}{ab}\\right)^2 \\geq 25[\/tex]<\/p>\n

[tex]\\qquad \\qquad \\dfrac{1}{4} \\left(1 + \\dfrac{1}{ab}\\right)^2 \\geq \\dfrac{25}{4}. \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,(IV)[\/tex]<\/p>\n

\n

Finalmente, a partir de [tex](III)[\/tex] e [tex](IV)[\/tex], conclu\u00edmos que<\/p>\n

[tex]\\qquad \\dfrac{\\left(a + \\frac{1}{a}\\right)^2 + \\left(b + \\frac{1}{b}\\right)^2}{2} \\geq \\dfrac{25}{4}[\/tex]<\/p>\n

donde<\/p>\n

[tex]\\qquad \\left(a + \\frac{1}{a}\\right)^2 + \\left(b + \\frac{1}{b}\\right)^2 \\geq \\dfrac{25}{2}[\/tex].<\/p>\n

\n

Solu\u00e7\u00e3o elaborada pelos Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n

\n\n\n
\"\"<\/td>\n\n

Aproveite o assunto do problema e d\u00ea uma passadinha na Sala para Leitura Algumas M\u00e9dias<\/a>.<\/span> <\/p>\n<\/td>\n<\/tr>\n<\/table>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Problema (Indicado a partir do 3\u00ba ano do E. M.) Prove as afirma\u00e7\u00f5es abaixo. (a) Para quaisquer [tex]x[\/tex] e [tex]y[\/tex] reais positivos temos [tex]\\left(\\dfrac{x+y}{2}\\right)^2 \\leq \\dfrac{x^2 + y^2}{2}[\/tex]; ocorrendo a igualdade se, e somente se, [tex]x=y[\/tex]. (b) Se [tex]a[\/tex] e [tex]b[\/tex] s\u00e3o reais positivos tais que [tex]a + b =1[\/tex], ent\u00e3o [tex]\\left(a + \\dfrac{1}{a}\\right)^2 + … <\/p>\n

Continue lendo<\/a><\/p>\n","protected":false},"author":179,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"open","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-42993","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/42993","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/179"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=42993"}],"version-history":[{"count":54,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/42993\/revisions"}],"predecessor-version":[{"id":185142,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/42993\/revisions\/185142"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=42993"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}