Notasi Sigma Apmo 1991


Misalkan a_1,a_2,a_3,...,a_n, b_1,b_2,b_3,...,b_n bilangan-bilangan real positif, sehingga

a_1+a_2+a_3+...+a_n=b_1+b_2+b_3+...+b_n .

Buktikan bahwa \sum \limits_{i=1}^n \frac {{a_i}^2}{a_i+b_i} \le \sum \limits_{i=1}^na_i

Solusi

Karena \sum \limits_{i=1}^na_i = \sum \limits_{i=1}^nb_i , maka

0 = \sum \limits_{i=1}^na_i - \sum \limits_{i=1}^nb_i

=\sum \limits_{i=1}^na_i-b_i

= \sum \limits_{i=1}^n \frac{{a_1}^2-{b_i}^2}{a_i+b_i}

= \sum \limits_{i=1}^n \frac{{a_i}^2}{a_i+b_i} - \sum \limits_{i=1}^n \frac{{b_i}^2}{a_i+b_i}

jelas bahwa \sum \limits_{i=1}^n \frac{{a_i}^2}{a_i+b_i} = \sum \limits_{i=1}^n \frac{{b_i}^2}{a_i+b_i}

Oleh karena itu diperoleh

\sum \limits_{i=1}^n \frac{{a_i}^2}{a_i+b_i}

=\frac{1}{2}( \sum \limits_{i=1}^n \frac{{a_i}^2}{a_i+b_i}- \sum \limits_{i=1}^n \frac{{b_i}^2}{a_i+b_i})

=\frac{1}{2} \sum \limits_{i=1}^n \frac{{a_1}^2+{b_i}^2}{a_i+b_i}

\le \frac{1}{2} \sum \limits_{i=1}^n \frac{{a_1}^2+2(a_ib_i)+{b_i}^2}{a_i+b_i}

= \frac{1}{2} \sum \limits_{i=1}^n \frac{(a_i+b_i)^2}{(a_i+b_i}

= \frac{1}{2} \sum \limits_{i=1}^n a_i+b_i

=\frac{1}{2} (\sum \limits_{i=1}^na_i+\sum \limits_{i=1}^nb_i)

=\frac{1}{2} (\sum \limits_{i=1}^na_i+\sum \limits_{i=1}^na_i)

=\sum \limits_{i=1}^na_i

About ardiantoarsadi

don't look for miracles it will come

Posted on Januari 23, 2010, in SOAL DAN SOLUSI and tagged , . Bookmark the permalink. Tinggalkan komentar.

Tinggalkan Balasan

Isikan data di bawah atau klik salah satu ikon untuk log in:

Logo WordPress.com

You are commenting using your WordPress.com account. Logout / Ubah )

Gambar Twitter

You are commenting using your Twitter account. Logout / Ubah )

Foto Facebook

You are commenting using your Facebook account. Logout / Ubah )

Foto Google+

You are commenting using your Google+ account. Logout / Ubah )

Connecting to %s

%d blogger menyukai ini: