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Doua inegalitati frumoase

 
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Mon May 24, 2010 11:01 pm    Post subject: Doua inegalitati frumoase Reply with quote

1. Fie a, b, c>0 cu a+b+c=1. Sa se arate ca:

\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+3(ab+bc+ca)\ge\frac{11}{2}.

2. Fie x, y, z>0 cu x+y+z=1. Sa se arate ca:

\frac{1+xy}{x+y}+\frac{1+yz}{y+z}+\frac{1+zx}{z+x}\ge 5.
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Brojbeanu Andi Gabriel, clasa XII-a
Colegiul National "Constantin Carabella" Targoviste
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Cristian Matei
Euclid


Joined: 15 Jul 2011
Posts: 18
Location: Sibiu

PostPosted: Fri Sep 09, 2011 3:04 pm    Post subject: Reply with quote

M-am gandit mult la prima insa nu am reusit sa o demonstrez.Daca cineva cunoaste solutia il rog frumos sa o posteze.
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Matei Cristian, clasa a VIII-a, Colegiul National ''Samuel von Brukenthal'' Sibiu
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Andi Brojbeanu
Newton


Joined: 22 Mar 2009
Posts: 383
Location: Targoviste (Dambovita)

PostPosted: Mon May 27, 2013 9:38 pm    Post subject: Reply with quote

1. Facem substitutiile a=\frac{x}{x+y+z}, b=\frac{y}{x+y+z}, c=\frac{z}{x+y+z} si p=x+y+z, q=xy+yz+zx, r=xyz. Atunci, inegalitatea este echivalenta cu:
\sum{\frac{1}{\frac{x}{x+y+z}+\frac{y}{x+y+z}}}+3\cdot \sum{\frac{xy}{(x+y+z)^2}}\ge \frac{11}{2} \Leftrightarrow \sum{x}\sum{\frac{1}{x+y}}+\frac{3q}{p^2}\ge \frac{11}{2} \Leftrightarrow p\cdot \frac{\sum{(z+x)(z+y)}}{\prod{(x+y)}}\ge \frac{11}{2}-\frac{3q}{p^2}\Leftrightarrow  \frac{\sum{x^2}+3\sum{xy}}{\sum{x}\sum{xy}-xyz}\ge \frac{11p^2-6q}{2p^3}
\Leftrightarrow  \frac{p^2+q}{pq-r}\ge \frac{11p^2-6q}{2p^3} \Leftrightarrow 2p^5+2p^3q\ge 11p^3q-11p^2r-6pq^2+6qr \Leftrightarrow 2p^5+11p^2r+6pq^2\ge 9p^3q+6qr .
Din inegalitatea lui Schur sub forma \sum{x^2(x-y)(x-z)}\ge 0 rezulta p^4+6pr+4q^2\ge 5p^2q.
Asadar, este suficient sa demonstram ca p^3q\ge 2pq^2+p^2r+6qr. Acest lucru rezulta imediat aplicand inegalitatile p^2\ge 3q, pq\ge 9r, p^3\ge 27r (rezulta
din inegalitatea mediilor): 2pq^2+p^2r+6qr\le 2pq\cdot \frac{p^2}{3}+p^2\cdot \frac{pq}{9}+6q\cdot \frac{p^3}{27}=p^3q(\frac{2}{3}+\frac{1}{9}+\frac{2}{9})=p^3q.
In fine, 2p^5+11p^2r+6pq^2=2p(p^4+6pr+4q^2)-p^2r-2pq^2\ge 2p\cdot 5p^2q-p^2r-2pq^2=9p^3q+(p^3q-2pq^2-p^2r)\ge 9p^3q+6qr, ceea ce
incheie demonstratia problemei.
2. Facand aceleasi substitutii, inegalitatea este echivalenta cu:
\sum{\frac{\frac{a+b+c}{a+b+c}}{\frac{a}{a+b+c}+\frac{b}{a+b+c}}}+\sum{\frac{\frac{ab}{(a+b+c)^2}}{\frac{a}{a+b+c}+\frac{b}{a+b+c}}}\ge 5 \Leftrightarrow \sum{\frac{a}{b+c}}+\frac{1}{a+b+c}\cdot \sum{\frac{ab}{a+b}}\ge 2 \Leftrightarrow \frac{\sum{a(a+b)(a+c)}}{\prod{(a+b)}}+\frac{\sum{ab(c+a)(c+b)}}{(a+b+c)\prod{(a+b)}}\ge 2\Leftrightarrow  \sum{a}\sum{a^3}+(\sum{a})^2\cdot
\cdot \sum{ab}+\sum{a}\cdot abc+(\sum{ab})^2\ge 2(a+b+c)(\sum{a}\sum{ab}-abc)\Leftrightarrow  p^2(p^2-3q)+3pr+p^2q+pr+q^2\ge 2p(pq-r)\Leftrightarrow p^4+q^2+6pr\ge 4p^2q.
Cum p^4+4q^2+6pr\ge 5p^2q si p^2\ge 3q, avem ca p^4+q^2+6pr=p^4+6pr+4q^2-3q^2\ge 5p^2q-q\cdot p^2=4p^2q, ceea ce incheie demonstratia.
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Brojbeanu Andi Gabriel, clasa XII-a
Colegiul National "Constantin Carabella" Targoviste
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