Olimpiade Zone : Pertidaksamaan

Berikut ini adalah kumpulan soal mengenai Pertidaksamaan tingkat olimpiade. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Tingkat Kesulitan :

Dasar SBMPTN Olimpiade

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No. 1

Buktikan bahwa jika x dan y adalah bilangan positif dan {xy=1}, maka {x+y\geq2}.

$\begin{array}{rl}{(\sqrt{x}-\sqrt{y})}^2& \ge 0\\ x-2\sqrt{xy}+y& \ge 0\\ x+y& \ge 2\sqrt{xy}\\ x+y& \ge 2\sqrt{1}\\ x+y& \ge 2(1)\\ x+y& \ge 2\end{array}$

Jadi, terbukti bahwa jika x dan y adalah bilangan positif dan xy=1, maka x+y\geq2.

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No. 2

Buktikan bahwa {\dfrac1{10\sqrt2}\lt\dfrac12\times\dfrac34\times\dfrac56\times\cdots\times\dfrac{99}{100}\lt\dfrac1{10}}

Misal {a=\dfrac12\times\dfrac34\times\dfrac56\times\cdots\times\dfrac{99}{100}}

$\begin{array}{rl}{a}^2& ={\displaystyle \frac{1^2}{2^2}}\times {\displaystyle \frac{3^2}{4^2}}\times {\displaystyle \frac{5^2}{6^2}}\times \cdots \times {\displaystyle \frac{{99}^2}{{100}^2}}\\ & ={\displaystyle \frac{1}{2\cdot 2}}\times {\displaystyle \frac{3^2}{4\cdot 4}}\times {\displaystyle \frac{5^2}{6\cdot 6}}\times \cdots \times {\displaystyle \frac{{99}^2}{100\cdot 100}}\\ & ={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3^2}{2\cdot 4}}\times {\displaystyle \frac{5^2}{4\cdot 6}}\times \cdots \times {\displaystyle \frac{{99}^2}{98\cdot 100}}\times {\displaystyle \frac{1}{100}}\end{array}$

Perhatikan bahwa {\dfrac{n^2}{(n-1)(n+1)}=\dfrac{n^2}{n^2-1}\gt1}

$\begin{array}{rl}{a}^2& ={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{3^2}{2\cdot 4}}\times {\displaystyle \frac{5^2}{4\cdot 6}}\times \cdots \times {\displaystyle \frac{{99}^2}{98\cdot 100}}\times {\displaystyle \frac{1}{100}}\\ & >{\displaystyle \frac{1}{2}}\times 1\times 1\times \cdots \times 1\times {\displaystyle \frac{1}{100}}\\ & >{\displaystyle \frac{1}{200}}\\ a& >{\displaystyle \frac{1}{10\sqrt{2}}}\\ {\displaystyle \frac{1}{10\sqrt{2}}}& <a& (1)\end{array}$

$\begin{array}{rl}{a}^2& ={\displaystyle \frac{1^2}{2^2}}\times {\displaystyle \frac{3^2}{4^2}}\times {\displaystyle \frac{5^2}{6^2}}\times \cdots \times {\displaystyle \frac{{99}^2}{{100}^2}}\\ & ={\displaystyle \frac{1}{2^2}}\times {\displaystyle \frac{3\cdot 3}{4^2}}\times {\displaystyle \frac{5\cdot 5}{6^2}}\times \cdots \times {\displaystyle \frac{99\cdot 99}{{100}^2}}\\ & ={\displaystyle \frac{1\cdot 3}{2^2}}\times {\displaystyle \frac{3\cdot 5}{4^2}}\times {\displaystyle \frac{5\cdot 7}{6^2}}\times \cdots \times {\displaystyle \frac{97\cdot 99}{{98}^2}}\times {\displaystyle \frac{99}{100}}\cdot {\displaystyle \frac{1}{100}}\end{array}$

Perhatikan bahwa {\dfrac{(n-1)(n+1)}{n^2}=\dfrac{n^2-1}{n^2}\lt1, dan \dfrac{99}{100}\lt1}

$\begin{array}{rl}{a}^2& ={\displaystyle \frac{1\cdot 3}{2^2}}\times {\displaystyle \frac{3\cdot 5}{4^2}}\times {\displaystyle \frac{5\cdot 7}{6^2}}\times \cdots \times {\displaystyle \frac{97\cdot 99}{{98}^2}}\times {\displaystyle \frac{99}{100}}\cdot {\displaystyle \frac{1}{100}}\\ & <1\times 1\times 1\times \cdots \times 1\times 1\cdot {\displaystyle \frac{1}{100}}\\ & <{\displaystyle \frac{1}{100}}\\ a& <{\displaystyle \frac{1}{10}}& (2)\end{array}$

Dari \color{red}{(1)} dan \color{red}{(2)} didapat,
{\dfrac1{10\sqrt2}\lt a\lt\dfrac1{10}}
{\dfrac1{10\sqrt2}\lt\dfrac12\times\dfrac34\times\dfrac56\times\cdots\times\dfrac{99}{100}\lt\dfrac1{10}}

Jadi, terbukti bahwa \dfrac1{10\sqrt2}\lt\dfrac12\times\dfrac34\times\dfrac56\times\cdots\times\dfrac{99}{100}\lt\dfrac1{10}.