Berikut ini adalah kumpulan soal mengenai Pertidaksamaan tingkat olimpiade. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
Buktikan bahwa jika
x dan
y adalah bilangan positif dan
{xy=1}, maka
{x+y\geq2}.
\(\begin{aligned}
\left(\sqrt{x}-\sqrt{y}\right)^2&\geq0\\
x-2\sqrt{xy}+y&\geq0\\
x+y&\geq2\sqrt{xy}\\
x+y&\geq2\sqrt{1}\\
x+y&\geq2(1)\\
x+y&\geq2
\end{aligned}\)
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}}
\(\begin{aligned}
a^2&=\dfrac{1^2}{2^2}\times\dfrac{3^2}{4^2}\times\dfrac{5^2}{6^2}\times\cdots\times\dfrac{99^2}{100^2}\\[8pt]
&=\dfrac1{2\cdot2}\times\dfrac{3^2}{4\cdot4}\times\dfrac{5^2}{6\cdot6}\times\cdots\times\dfrac{99^2}{100\cdot100}\\[8pt]
&=\dfrac12\times\dfrac{3^2}{2\cdot4}\times\dfrac{5^2}{4\cdot6}\times\cdots\times\dfrac{99^2}{98\cdot100}\times\dfrac1{100}
\end{aligned}\)
Perhatikan bahwa {\dfrac{n^2}{(n-1)(n+1)}=\dfrac{n^2}{n^2-1}\gt1}
\(\begin{aligned}
a^2&=\dfrac12\times\dfrac{3^2}{2\cdot4}\times\dfrac{5^2}{4\cdot6}\times\cdots\times\dfrac{99^2}{98\cdot100}\times\dfrac1{100}\\[8pt]
&\gt\dfrac12\times1\times1\times\cdots\times1\times\dfrac1{100}\\[8pt]
&\gt\dfrac1{200}\\[8pt]
a&\gt\dfrac1{10\sqrt2}\\[8pt]
\dfrac1{10\sqrt2}&\lt a\qquad&\color{red}{(1)}
\end{aligned}\)
\(\begin{aligned}
a^2&=\dfrac{1^2}{2^2}\times\dfrac{3^2}{4^2}\times\dfrac{5^2}{6^2}\times\cdots\times\dfrac{99^2}{100^2}\\[8pt]
&=\dfrac1{2^2}\times\dfrac{3\cdot3}{4^2}\times\dfrac{5\cdot5}{6^2}\times\cdots\times\dfrac{99\cdot99}{100^2}\\[8pt]
&=\dfrac{1\cdot3}{2^2}\times\dfrac{3\cdot5}{4^2}\times\dfrac{5\cdot7}{6^2}\times\cdots\times\dfrac{97\cdot99}{98^2}\times\dfrac{99}{100}\cdot\dfrac1{100}
\end{aligned}\)
Perhatikan bahwa {\dfrac{(n-1)(n+1)}{n^2}=\dfrac{n^2-1}{n^2}\lt1, dan \dfrac{99}{100}\lt1}
\(\begin{aligned}
a^2&=\dfrac{1\cdot3}{2^2}\times\dfrac{3\cdot5}{4^2}\times\dfrac{5\cdot7}{6^2}\times\cdots\times\dfrac{97\cdot99}{98^2}\times\dfrac{99}{100}\cdot\dfrac1{100}\\[8pt]
&\lt1\times1\times1\times\cdots\times1\times1\cdot\dfrac1{100}\\[8pt]
&\lt\dfrac1{100}\\[8pt]
a&\lt\dfrac1{10}\qquad&\color{red}{(2)}
\end{aligned}\)
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}}
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