Contoh Soal 12 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 56.& (\mathbf{\text{OSK 2018}})\\ & \text{Diketahui bilangan real}x\text{dan}y\\ & \text{yang memenuhi}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Nilai minimum}{\displaystyle \frac{x}{2y-x}+\frac{2y}{2x-y}\text{adalah}....}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Misal}t={\displaystyle \frac{x}{y}}\\ & \text{Misalkan juga}f(t)={\displaystyle \frac{x}{2y-x}+{\displaystyle \frac{2y}{2x-y}}}\\ & \text{maka}f(t)={\displaystyle \frac{2t^2-3t+4}{-2t^2+5t-2}}\\ & \bullet \text{Agar minimum, maka}{f}^{\prime }(t)=0\\ & \text{Sehingga}\\ & f^{\prime }(t)=4t^2+8t-14=0\\ & \iff t_{1,2}=-1\pm {\displaystyle \frac{3}{2}\sqrt{2}}\\ & \text{Pilih yang positif, yaitu}t=-1+{\displaystyle \frac{3}{2}\sqrt{2}}\\ & \bullet \text{Dengan proses substistusi harga}t\\ & \text{di atas, maka akan didapatkan }\\ & \text{nilai}f(t)=1+{\displaystyle \frac{4}{3}\sqrt{2}}\end{array}\\ & \begin{array}{rl}& \text{Alternatif 2}\\ & \text{Menurut bentuk}{\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{jelas bahwa baik}2y-x\text{dan}2x-y\\ & \text{keduanya}\mathbf{\text{positif}}\\ & \text{Lihat tabel berikut}\\ & \begin{array}{|ccc|}\hline \text{Bentuk}& \text{Pengecekan 1}& \text{Pengecekan 2}\\ \begin{array}{rl}& {\displaystyle \frac{1}{2}<\frac{x}{y}<2}\\ & \text{Jelas bahwa}\\ & x,y\ne 0\\ & \\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat}(\times y)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(y)<\frac{x}{y}(y)<2(y)}\\ & \iff {\displaystyle \frac{1}{2}y<x<2y}\\ & \text{Jelas bahwa}\\ & 2y-x>0\\ & \\ & \\ & \end{array}& \begin{array}{rl}& \text{Saat dibali posisinya}\\ & {\displaystyle \frac{1}{2}<\frac{y}{x}<2}\\ & \text{Saat}(\times x)\\ & \text{Yaitu}:\\ & {\displaystyle \frac{1}{2}(x)<\frac{y}{x}(x)<2(x)}\\ & \iff {\displaystyle \frac{1}{2}x<y<2x}\\ & \text{Jelas bahwa}\\ & 2x-y>0\end{array}\\ \hline\end{array}\\ & \text{Saat masing-masing}\\ & \bullet {\displaystyle \frac{x}{2y-x}={\displaystyle \frac{1}{3}+\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)\text{dan}}}\\ & \bullet {\displaystyle \frac{2y}{2x-y}={\displaystyle \frac{2}{3}+\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)}}\\ & \text{Dengan ketaksamaan AM-GM diperoleh}\\ & \begin{array}{rl}{\displaystyle \frac{x}{2y-x}+{\displaystyle \frac{2y}{2x-y}}}& =1+\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)+\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)\\ & \ge 1+2\sqrt{\frac{2}{3}\left({\displaystyle \frac{2x-y}{2y-x}}\right)\frac{4}{3}\left({\displaystyle \frac{2y-x}{2x-y}}\right)}\\ & =1+2\sqrt{{\displaystyle \frac{8}{9}}}\\ & =1+2\left({\displaystyle \frac{2}{3}}\right)\sqrt{2}\\ & =1+{\displaystyle \frac{4}{3}\sqrt{2}}\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 57.& \text{Diketahui}a,b\text{bilangan real positif}\\ & \text{dan}a+b=1.\text{Tunjukkan bahwa}\\ & {(a+{\displaystyle \frac{1}{a}})}^2+{(b+{\displaystyle \frac{1}{b}})}^2\ge {\displaystyle \frac{25}{2}}\\ \\ & \mathbf{\text{Bukti}}\end{array}$.

$\begin{array}{rl}& \text{Dengan AM-GM kita akan mendapatkan}\\ & 1=a+b\ge 2\sqrt{ab}\iff {\displaystyle \frac{1}{2}\ge \sqrt{ab}\iff 2\le {\displaystyle \frac{1}{\sqrt{ab}}}}\\ & \iff {\displaystyle \frac{1}{\sqrt{ab}}\ge 2\iff {\displaystyle \frac{1}{ab}\ge 4}}\\ & \text{Perhatikan soal, dengan CS-Engel}\text{akan}\\ & \text{didapatkan}\\ & \begin{array}{rl}& (1+1)\left({\displaystyle \frac{{(a+\frac{1}{a})}^2}{1}+{\displaystyle \frac{{(b+\frac{1}{b})}^2}{1}}}\right)\ge {(a+\frac{1}{a}+b+\frac{1}{b})}^2\\ & \iff 2({\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2)\ge {(a+b+\frac{1}{a}+\frac{1}{b})}^2\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{a+b}{ab}})}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}{(1+{\displaystyle \frac{1}{ab}})}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}(1+4{)}^2}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{1}{2}\times 25}\\ & \iff {\left({\displaystyle a+\frac{1}{a}}\right)}^2+{(b+\frac{1}{b})}^2\ge {\displaystyle \frac{25}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 58.& \text{Jika}a,b\text{bilangan positif, buktikan}\\ & \text{a}.2(a^2+b^2)\ge (a+b{)}^2\\ & \text{b}.4(a^3+b^3)\ge (a+b{)}^3\\ & \text{c}.8(a^4+b^4)\ge (a+b{)}^4\\ & \text{d}.16(a^5+b^5)\ge (a+b{)}^5\\ & \text{e}.32(a^6+b^6)\ge (a+b{)}^6\\ & \text{f}.64(a^7+b^7)\ge (a+b{)}^7\\ & \text{g}.128(a^8+b^8)\ge (a+b{)}^8\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Akan ditunjukkan bukti poin 6.c saja}\\ & \text{untuk poin yang lain, silahkan pembaca}\\ & \text{sekalian untuk dibuktikan sendiri sebagai}\\ & \text{bahan latihan mandiri}.\\ & \text{Adapun bukti poin 6.c adalah sebagaimana}\\ & \text{berikut ini}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan CS-Engel}\text{akan}\\ & \text{didapatkan}\\ & \bullet (1+1)(a^4+b^4)\ge (a^2+b^2{)}^2\\ & \iff 2(a^4+b^4)\ge (a^2+b^2{)}^2..........(1)\\ & \bullet (1+1)(a^2+b^2)\ge (a+b{)}^2\\ & \iff 2(a^2+b^2)\ge {\displaystyle (a+b{)}^2,(\text{kuadratkan})}\\ & \iff 4{\displaystyle {(a^2+b^2)}^2\ge (a+b{)}^4}\\ & \iff {(a^2+b^2)}^2\ge {\displaystyle \frac{(a+b{)}^4}{4}...........(2)}\\ & \text{Dari (1) dan (2) didapatkan hubungan}\\ & 2(a^4+b^4)\ge {(a^2+b^2)}^2\ge {\displaystyle \frac{(a+b{)}^4}{4}}\\ & \iff 8(a^4+b^4)\ge 4{(a^2+b^2)}^2\ge (a+b{)}^4\\ & \iff 8(a^4+b^4)\ge (a+b{)}^4◼\end{array}\end{array}$.

$\begin{array}{l}\\ 59.& \text{Jika}x,y,z\text{adalah bilangan real positif}\\ & \text{dengan}{x}^2+y^2+z^2=27.\text{Tunjukkan}\\ & \text{bahwa}{x}^3+y^3+z^3\ge 81\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Pada contoh soal no.2 terdapat}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{ganti mengganti}a=b=c=1,\text{maka menjadi}\\ & (x+y+z{)}^2\le (1^2+1^2+1^2)(x^2+y^2+z^2)\\ & \iff (x+y+z{)}^2\le (3)(x^2+y^2+z^2)........(1)\\ & \text{Selanjutnya dengan mengganti dengan}\\ & x^{{.}^{\frac{3}{2}}},y^{{.}^{\frac{3}{2}}},z^{{.}^{\frac{3}{2}}}\text{dan}{x}^{{.}^{\frac{1}{2}}},y^{{.}^{\frac{1}{2}}},z^{{.}^{\frac{1}{2}}},\text{pada}\\ & (ax+by+cz{)}^2\le (a^2+b^2+c^2)(x^2+y^2+z^2)\\ & \text{Kita akan dapatkan}\\ & (x^2+y^2+z^2)\le (x^3+y^3+z^3)(x+y+z)........(2)\\ & \text{Jika masing-masing ruas dikuadratkan, maka}\\ & (x^2+y^2+z^2{)}^4\le (x^3+y^3+z^3{)}^2(x+y+z{)}^2\\ & \iff (x^2+y^2+z^2{)}^4\le 3(x^3+y^3+z^3{)}^2(x^2+y^2+z^2)\\ & \iff (x^2+y^2+z^2{)}^3\le 3(x^3+y^3+z^3{)}^2\\ & \iff (27{)}^3\le 3(x^3+y^3+z^3{)}^2\\ & \iff 3^9\le 3(x^3+y^3+z^3{)}^2\\ & \iff 3^8\le (x^3+y^3+z^3{)}^2\\ & \iff (x^3+y^3+z^3{)}^2\ge 3^8\\ & \iff (x^3+y^3+z^3)\ge 3^{{.}^{\frac{8}{2}}}\\ & \iff (x^3+y^3+z^3)\ge 3^4=81◼\end{array}$.

$\begin{array}{l}\\ 60.& \text{Untuk}a,b,c,d\text{adalah bilangan real }\\ & \text{positif, tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge {\displaystyle \frac{64}{a+b+c+d}}}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Dengan ketaksamaan CS-Engel}\\ & (a+b+c+d)\left({\displaystyle \frac{1^2}{a}+{\displaystyle \frac{1^2}{b}+{\displaystyle \frac{2^2}{c}+{\displaystyle \frac{4^2}{d}}}}}\right)\ge (1+1+2+4{)}^2\\ & \iff {\displaystyle \frac{1^2}{a}+{\displaystyle \frac{1^2}{b}+{\displaystyle \frac{2^2}{c}+{\displaystyle \frac{4^2}{d}\ge {\displaystyle \frac{(1+1+2+4{)}^2}{a+b+c+d}}}}}}\\ & \iff {\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{4}{c}+\frac{16}{d}\ge {\displaystyle \frac{64}{a+b+c+d}◼}}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Muslim, M.S. 2020. Kumpulan Soal dan Pembahasan Olimpiade Matematika SMA Tahun 2007-2019 Tingkat Kota/Kabupaten. Bandung: YRAMA WIDYA.
2. Widodo, T. 2018. Booklet OSN SMA 2018: Soal dan Solusi OSK, OSP, OSN SMA Bidang Matematika.
3. Young, B. 2009. Seri Buku Olimpiade Matematika Strategi Menyelesaikan Soal-Soal Olimpiade Matematika: Ketaksamaan (Inequality). Bandung: PAKAR RAYA.