Ketaksamaan Keren


Misalkan a,b, dan c adalah bilangan real positif. Buktikan bahwa

\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}


SOLUSI

\frac{1}{abc}-(\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc})

=\frac{a^6b^3+a^3b^6+b^6c^3+b^3c^6+a^6c^3+a^3c^6-2a^5b^2c^2-2a^2b^5c^2-2a^2b^2c^5}{abc(a^3+b^3+abc)(b^3+c^3+abc)(a^3+c^3+abc)}

=\frac{(a^3+b^3+c^3)(a^3b^3+b^3c^3+c^3a^3-2a^2b^2c^2)}{abc(a^3+b^3+abc)(b^3+c^3+abc)(a^3+c^3+abc)}

Menurut ketaksamaan AM-GM diperoleh

a^3b^3+b^3c^3+c^3a^3\ge 3\sqrt[3]{(a^3b^3)(b^3c^3)(c^3a^3)}=3a^2b^2c^2

Sehingga

a^3b^3+b^3c^3+c^3a^3-2a^2b^2c^2\ge3a^2b^2c^2-2a^2b^2c^2=a^2b^2c^2\ >0

Akibatnya

\frac{1}{abc}-(\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc}) >0

Hal ini equivalen dengan

\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{a^3+c^3+abc} \le \frac{1}{abc}

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.

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