Lanjutan Ketaksamaan Schur

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1. Penyederhanaan dengan pola siklik dan simetri

$.\begin{array}{rl}& \text{Mengenal penulisan pola}\mathbf{\text{Siklik dan Simetri}}\\ & \text{Misal untuk}n=3,\text{pada penulisan unsur}\\ & x,y,\text{dan}z,\text{maka}\\ & \begin{array}{|ll|}\hline \mathbf{\text{Pola Siklik}}& \mathbf{\text{Pola Simetri}}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^2=x^2+y^2+z^2}\\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}{x}^2}& =x^2+x^2\\ & +y^2+y^2\\ \\ & +z^2+z^2\\ \\ & =2(x^2+y^2+z^2)\end{array}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^3=x^3+y^3+z^3}\end{array}& \begin{array}{r}{\displaystyle \sum _{\text{sym}}^{.}{x}^3=2(x^3+y^3+z^3)}\end{array}\\ \begin{array}{r}{\displaystyle \sum _{\text{siklik}}^{.}{x}^{2}y=x^{2}y+y^{2}z+z^{2}x}\\ & \\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}{x}^{2}y}& =x^{2}y+x^{2}z\\ & +y^{2}x+y^{2}z\\ \\ & +z^{2}x+z^{2}y\end{array}\\ \begin{array}{rl}{\displaystyle \sum _{\text{siklik}}^{.}xyz}& =xyz+yzx+zxy\\ & =3xyz\end{array}& \begin{array}{rl}{\displaystyle \sum _{\text{sym}}^{.}xyz}& =xyz+xzy+\cdots \\ & =6xyz\end{array}\\ \hline\end{array}\end{array}$

2. Bentuk ketaksamaan berdasar nilai r

Masih ingat kita pada ketaksamaan Schur saat $r=1$, yaitu,:

$\begin{array}{rl}& a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)\\ & =a^3+b^3+c^3-(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b)+3abc\\ & ={\displaystyle \sum _{\text{sym}}^{.}(a^3-2a^{2}b+abc)\ge 0}\end{array}$.

Selanjutnya saat  $r=1$, kita bisa mendaptkan

$\begin{array}{rl}1.& a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a)\\ 2.& abc\ge (a+b-c)(b+c-a)(c+a-b)\\ 3.& (a+b+c{)}^3+9abc\ge 4(a+b+c)(ab+bc+ca)\end{array}$.

Dan saat  $r=2$, kita bisa mendaptkan

$\begin{array}{rl}{a}^4+& b^4+c^4+abc(a+b+c)\ge ab(a^2+b^2)+bc(b^2+c^2)+ca(c^2+a^2)\end{array}$.

3. Beberapa formulasi bantu

$\begin{array}{rl}1.& (a+b+c)(a^2+b^2+c^2)\\ & =a^3+b^3+c^3+a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)\\ 2.& (a+b+c)(a^2+b^2+c^2)\\ & =a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 3.& (a+b+c)(ab+ac+bc)\\ & =3abc+ab(a+b)+ac(a+c)+bc(b+c)\\ 4.& (a+b+c)(ab+ac+bc)\\ & =3abc+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 5.& (a+b+c{)}^3+3abc\\ & =a^3+b^3+c^3+3(a+b+c)(ab+ac+bc)\\ 6.& (a+b)(a+c)(b+c)\\ & =2abc++a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 7.& (a+b-c)(a+c-b)(b+c-a)\\ & =-2abc-(a^3+b^3+c^3)+a^2(b+c)+b^2(a+c)+c^2(a+b)\\ 8.& (a-b)(b-c)(c-a)\\ & =a^2(c-b)+b^2(a-c)+c^2(b-a)\end{array}$.

4. Penyederhanaan ketaksamaan metode pqr

$\begin{array}{|ll|}\hline .\mathbf{\text{Kesamaan}}& \mathbf{\text{Ketaksamaan}}\\ \begin{array}{rl}1.& {\displaystyle \sum _{sik}^{.}{x}^2=p^2-2q}\\ 2.& {\displaystyle \sum _{sik}^{.}{x}^3=p(p^2-3q)+3r}\\ 3.& \sum _{sik}^{.}{x}^2{y}^2=q^2-2pr\\ 4.& \prod (x+y)=pq-r\\ 5.& \sum _{sik}^{.}xy(x+y)=pq-3r\\ 6.& \sum _{sik}^{.}{x}^2(y+z)=pq-3r\\ 7.& \prod (1+x)=1+p+q+r\\ & \\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}1.& pq\ge 9r\\ 2.& p^2\ge 3q\\ 3.& q^2\ge 3pr\\ 4.& p^3\ge 27r\\ 5.& q^3\ge 27r^2\\ 6.& p^{3}r\ge q^3\\ 7.& p^3+9r\ge 4pq\\ 8.& 2p^3+9r\ge 7pq\\ 9.& 2p^3+27r\ge 9pq\\ 10.& 2p^3+9r^2\ge 7pqr\\ 11.& q^3+9r^2\ge 4pqr\\ 12.& 2q^3+27r^2\ge 9pqr\\ 13.& p^4+3q^2\ge 4p^{2}q\\ 14.& p^4+4q^2+6pr\ge 5p^{2}q\\ 15.& p^{2}q+3pr\ge 4q^2\\ 16.& p{q}^2\ge 2p^{2}r+3qr\\ 17.& p^2{q}^2+12r^2\ge 4p^{3}r+pqr\end{array}\\ \hline\end{array}$.

5. Ketaksamaan Schur bentuk pqr

Perhatikan poin 2 di atas saat $r=1$, kitaa akan medapatkan bentuk berikut ini:

$\begin{array}{rl}& (a+b+c{)}^3+9abc\ge 4(a+b+c)(ab+bc+ca)\\ & \iff p^3+9r\ge 4pq\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1& \text{Untuk}a,b,c\text{tak negatif, tunjukkan bahwa}\\ & (a+b+c)(ab+ac+bc)\ge 9abc\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Untuk pembuktian pernyataan di atas}\\ & \text{dengan AM-GM kita memiliki}\\ & \bullet a+b+c\ge 3\sqrt[3]{abc}\\ & \bullet ab+ac+bc\ge 3\sqrt[3]{(abc{)}^2}\\ & \text{maka hasil dari}\\ & (a+b+c)(ab+ac+bc)\ge 3\sqrt{abc}.3\sqrt[3]{(abc{)}^2}\\ & \iff pq\ge 3\sqrt[3]{r}.3\sqrt[3]{r^2}\\ & \iff pq\ge 9\sqrt[3]{r^3}\\ & \iff pq\ge 9r◼\end{array}$.

$\begin{array}{l}\\ 2& \text{Untuk}a,b,c\text{tak negatif, tunjukkan bahwa}\\ & (a+b+c{)}^2\ge 3(ab+ac+bc)\\ \\ & \mathbf{\text{Bukti}}\\ & \mathbf{\text{Alternatif 1}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Untuk pembuktian pernyataan di atas}\\ & \text{dengan AM-GM kita memiliki}\\ & \bullet a^2+b^2+c^2\ge ab+ac+bc\\ & \text{Dan juga sebuah kesamaan}\\ & \bullet a^2+b^2+c^2=(a+b+c{)}^2-2(ab+ac+bc)\\ & \text{maka dari kedua bentuk di atas kita akan}\\ & \text{dapatkan bentuk}\\ & a^2+b^2+c^2\ge ab+ac+bc\\ & \iff (a+b+c{)}^2-2(ab+ac+bc)\ge ab+ac+bc\\ & \iff (a+b+c{)}^2\ge 3(ab+ac+bc)\\ & \iff p^2\ge 3q◼\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & \text{Telah kita ketahui bahwa}\\ & (a+b+c{)}^2=a^2+b^2+c^2+2(ab+ac+bc)\\ & \text{Dengan ketaksamaan}\mathbf{\text{renata}}\text{kita akan}\\ & \text{dapatkan bentuk}\\ & (a+b+c{)}^2\ge ab+bc+ca+2(ab+ac+bc)\\ & \iff (a+b+c{)}^2\ge 3(ab+ac+bc)\\ & \iff p^2\ge 3q◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Untuk}a,b,c\text{tak negatif, tunjukkan}\\ & \text{bahwa}(ab+ac+bc{)}^2\ge 3abc(a+b+c)\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Perhatikan kesamaan berikut}\\ & \begin{array}{rl}& (ab+ac+bc{)}^2=a^2{b}^2+a^2{c}^2+b^2{c}^2+2abc(a+b+c)\\ & \text{Dengan AM-GM dan GM-AM kita dapatkan}\\ & \iff (ab+ac+bc{)}^2\ge 3\sqrt[3]{a^4{b}^4{c}^4}+2abc(a+b+c)\\ & \iff (ab+ac+bc{)}^2\ge 3abc\sqrt[3]{abc}+2abc(a+b+c)\\ & \iff (ab+ac+bc{)}^2\ge abc(a+b+c)+2abc(a+b+c)\\ & \iff (ab+ac+bc{)}^2\ge 3abc(a+b+c)\\ & \iff q^2\ge 3pr◼\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Untuk}a,b,c\text{tak negatif, tunjukkan bahwa}\\ & (a+b+c{)}^3\ge 27abc\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Untuk pembuktian pernyataan di atas}\\ & \text{dengan AM-GM kita memiliki}\\ & \bullet a+b+c\ge 3\sqrt[3]{abc}\\ & \text{Masing-masing ruas pangkatkan 3, maka}\\ & (a+b+c{)}^3\ge 27abc\iff p^3\ge 27r◼\end{array}$.

$\begin{array}{l}\\ 5.& \text{Untuk}a,b,c\text{tak negatif, tunjukkan bahwa}\\ & (a+b+c{)}^3\ge 27abc\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Misalkan}\\ & p=a+b+c,q=ab+ac+bc,\\ & \text{dan}r=abc\\ & \text{Untuk pembuktian pernyataan di atas}\\ & \text{dengan AM-GM kita memiliki}\\ & \bullet ab+ac+bc\ge 3\sqrt[3]{(abc{)}^2}\\ & \text{Masing-masing ruas pangkatkan 3, maka}\\ & (ab+ac+bc{)}^3\ge 27(abc{)}^2\iff q^3\ge 27r^2◼\end{array}$.

$\begin{array}{l}\\ 6.& \text{Misalkan}x,y,x\text{bilangan real positif dengan}\\ & \text{tunjukkan bahwa}\\ & a^2+b^2+c^2+2abc+1\ge 2(ab+acb+c)\\ & (\mathbf{\text{Darij Grinberg}})\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\text{dan}\\ & \text{dilanjutkan dengan}\mathbf{\text{ketaksamaan Schur}}\\ & \text{serta menggesernya ke ruas kiri, maka}\\ & a^2+b^2+c^2+2abc+1-2(ab+ac+bc)\\ & \ge a^2+b^2+c^2+3(abc{)}^{{}^{\frac{2}{3}}}+1\ge 2(ab+ac+bc)\\ & \ge {((a{)}^{{}^{\frac{2}{3}}})}^3+{((b{)}^{{}^{\frac{2}{3}}})}^3+{((c{)}^{{}^{\frac{2}{3}}})}^3+3(abc{)}^{{}^{\frac{2}{3}}}-2(ab+ac+bc)\\ & \ge a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+b^{{.}^{\frac{2}{3}}})+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(b^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})-2(ab+ac+bc)\\ & \ge a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+b^{{.}^{\frac{2}{3}}})+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(a^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}(b^{{.}^{\frac{2}{3}}}+c^{{.}^{\frac{2}{3}}})-2(ab+ac+bc)\\ & =a^{{.}^{\frac{2}{3}}}{b}^{{.}^{\frac{2}{3}}}{(a^{{.}^{\frac{2}{3}}}-b^{{.}^{\frac{2}{3}}})}^2+a^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}{(a^{{.}^{\frac{2}{3}}}-c^{{.}^{\frac{2}{3}}})}^2\\ & +b^{{.}^{\frac{2}{3}}}{c}^{{.}^{\frac{2}{3}}}{(b^{{.}^{\frac{2}{3}}}-c^{{.}^{\frac{2}{3}}})}^2\ge 0◼\end{array}$.

$\begin{array}{l}\\ 7.& (\mathbf{\text{APMO 2004}})\\ & \text{Misalkan}x,y,x\text{bilangan real positif dengan}\\ & \text{tunjukkan bahwa}\\ & (x^2+2)(y^2+2)(z^2+2)\ge 9(xy+yz+zx)\\ \\ & \mathbf{\text{Bukti}}\\ & \mathbf{\text{Alternatif 1}}\\ & \text{Dengan menjabarkan akan didapatkan}\\ & x^2{y}^2{z}^2+2{\displaystyle \sum _{\text{siklik}}^{.}{x}^2{y}^2+4{\displaystyle \sum _{\text{siklik}}^{.}{x}^2+8\ge 9{\displaystyle \sum _{\text{siklik}}^{.}xy}}}\\ & \text{Perhatikan bahwa}\\ & \bullet 2{\displaystyle \sum _{\text{siklik}}^{.}{x}^2{y}^2-4{\displaystyle \sum _{\text{siklik}}^{.}xy+6=2{\displaystyle \sum _{\text{siklik}}^{.}(xy-1{)}^2\ge 0}}}\\ & \bullet {\displaystyle \sum _{\text{siklik}}^{.}{x}^2\ge {\displaystyle \sum _{\text{siklik}}^{.}xy\text{atau}{x}^2+y^2+z^2\ge xy+xz+yz}}\\ & \text{Kita cukup membuktikan bahwa}\\ & x^2{y}^2{z}^2+{\displaystyle \sum _{\text{siklik}}^{.}{x}^2+2\ge 2{\displaystyle \sum _{\text{siklik}}^{.}xy}}\\ & \iff x^2{y}^2{z}^2+2\ge {\displaystyle \sum _{\text{siklik}}^{.}xy}\\ & \begin{array}{rl}& \text{Untuk}a,b,c\text{bilangan real positif},\\ & \mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1\\ & \text{memberikan}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{a}^3+3abc\ge {\displaystyle \sum _{\text{siklik}}^{.}{a}^{2}b+{\displaystyle \sum _{\text{siklik}}^{.}a{b}^2}}}\\ & =ab(a+b)+bc(b+c)+ca(c+a)\\ & \text{Dengan}\mathbf{\text{Ketaksamaan AM-GM}}\text{didapakan}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{a}^3+3abc\ge 2{\displaystyle \sum _{\text{siklik}}^{.}(ab{)}^{\frac{3}{2}}}}\\ & \text{Pilih}a=x^{\frac{2}{3}},b=y^{\frac{2}{3}},c=z^{\frac{2}{3}},\text{maka didapatkan}\\ & (x^{\frac{2}{3}}{)}^3+(y^{\frac{2}{3}}{)}^3+(z^{\frac{2}{3}}{)}^3+3(xyz{)}^{\frac{2}{3}}\ge 2(xy+yz+zx)\\ & \text{Selanjutnya kita selesaikan ini},x^2{y}^2{z}^2+2\ge {\displaystyle \sum _{\text{siklik}}^{.}xy}\\ & \iff x^2{y}^2{z}^2+2\ge 3(xyz{)}^{\frac{2}{3}}\\ & \text{Misalkan}(xyz{)}^{\frac{2}{3}}=t,\text{maka}\\ & t^3+2\ge 3t\iff t^3-3t+2\ge 0\\ & (t-1{)}^2(t+2)\ge 0\text{adalah hal benar}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & x^2{y}^2{z}^2+2{\displaystyle \sum _{\text{siklik}}^{.}{x}^2{y}^2+4{\displaystyle \sum _{\text{siklik}}^{.}{x}^2+8\ge 9{\displaystyle \sum _{\text{siklik}}^{.}xy}}}\\ & \text{atau dalam bentuk utuhnya, yaitu}\\ & x^2{y}^2{z}^2+2(x^2{y}^2+x^2{z}^2+y^2{z}^2)+4(x^2+y^2+z^2)+8\ge 9(xy+xz+yz)\\ & \text{Sekarang kira uraikan satu persatu bagian}\\ & \bullet x^2{y}^2{z}^2+1+1\ge 3\sqrt[3]{(xyz{)}^2}\ge {\displaystyle \frac{9abc}{a+b+c}=\frac{9r}{p}}\\ & \text{ingat bahwa}\mathbf{\text{jika ada}}{\displaystyle \frac{9r}{p}\ge 4q-p^2}\\ & =4(xy+xz+yz)-(x+y+z{)}^2\\ & \text{adalah}\mathbf{\text{ketaksamaan Schur saat}}r=1\\ & \bullet x^2{y}^2+1+x^2{z}^2+1+y^2{z}^2+1\ge 2(xy+xz+yz)\\ & \bullet x^2+y^2+z^2\ge xy+xz+yz\\ & \text{keduanya didapat dengan ketaksamaan}\mathbf{\text{AM-GM}}\\ & x^2{y}^2{z}^2+2(x^2{y}^2+x^2{z}^2+y^2{z}^2)+4(x^2+y^2+z^2)+8\\ & =x^2{y}^2{z}^2+2+2(x^2{y}^2+x^2{z}^2+y^2{z}^2+3)+4(x^2+y^2+z^2)\\ & \ge 4(xy+xz+yz)-(x+y+z{)}^2+4(xy+xz+yz)+4(xy+xz+yz)\\ & \ge 12(xy+xz+yz)-(x+y+z{)}^2\\ & \ge 12(xy+xz+yz)-3(xy+xz+yz)\\ & =9(xy+xz+yz)◼\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Alternatif 3}}\\ & (x^2+2)(y^2+2)(z^2+2)-9(xy+yz+zx)\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\\ & =x^2{y}^2{z}^2+2(x^2{y}^2+x^2{z}^2+y^2{z}^2)\\ & +4(x^2+y^2+z^2)+8-9(xy+xz+yz)\\ & =4(x^2+y^2+z^2)+2((x^2{y}^2+1)+(x^2{z}^2+1)+(y^2{z}^2+1))\\ & +(x^2{y}^2{z}^2+1)+1-9(xy+xz+yz)\\ & \ge 4(x^2+y^2+z^2)+4(xy+xz+yz)\\ & +2xyz+1-9(xy+xz+yz)\\ & =(x^2+y^2+z^2)+3(x^2+y^2+z^2)\\ & +2xyz+1-5(x^2+y^2+z^2)\\ & \ge x^2+y^2+z^2+3(xy+xz+yz)\\ & +2xyz-5(xy+xz+yz)\\ & =x^2+y^2+z^2+2xyz+1-2(xy+xz+yz)\ge 0\\ & \text{adalah benar dengan bukti ada pada}\\ & \text{nomor soal sebelumnya}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Misalkan}x,y,x\text{bilangan real positif dengan}\\ & \text{tunjukkan bahwa}\\ & 2(a^2+b^2+c^2)+abc+8\ge 5(a+b+c)\\ & (\mathbf{\text{Tran Nam Dung}})\\ \\ & \mathbf{\text{Bukti}}\\ & \text{Dengan}\mathbf{\text{ketaksamaan AM-GM}}\text{dan}\\ & \text{menggeser ke ruas kiri dan masing-masing}\\ & \text{serta mengalikan semunya dengan 6, maka}\\ & 12(a^2+b^2+c^2)+6abc+48-30(a+b+c)\\ & =12(a^2+b^2+c^2)+3(2abc+1)+45-5.2.3(a+b+c)\\ & \ge 2(a^2+b^2+c^2)+9\sqrt[3]{(abc{)}^2}+45-5((a+b+c{)}^2+9)\\ & =12(a^2+b^2+c^2)+{\displaystyle \frac{9abc}{\sqrt[3]{abc}}-5((a^2+b^2+c^2)+2(ab+ac+bc))}\\ & =7(a^2+b^2+c^2)+{\displaystyle \frac{9abc}{\sqrt[3]{abc}}-10(ab+ac+bc)}\\ & \ge 7(a^2+b^2+c^2)+{\displaystyle \frac{27abc}{a+b+c}-10(ab+ac+bc)}\\ & \begin{array}{rl}& \text{dengan}\mathbf{\text{ketaksamaan Schur}},\text{yaitu}:\\ & p^3+9r\ge 4pq\iff {\displaystyle \frac{9r}{p}\ge 4q-p^2}\\ & \text{maka ketaksamaan akan menjadi}\\ & \ge 7(a^2+b^2+c^2)+3(4q-p^2)-10q\\ & \ge 7(a^2+b^2+c^2)+2q-3p^2\\ & =7(a^2+b^2+c^2)+2(ab+ac+bc)-3(a+b+c{)}^2\\ & =7(a^2+b^2+c^2)+2q-3((a^2+b^2+c^2)+2q)\\ & =4(a^2+b^2+c^2)+2q-6q\\ & =4(a^2+b^2+c^2)-4q\\ & =4(a^2+b^2+c^2)-4(ab+ac+bc)\\ & =4(a^2+b^2+c^2-ab-ac-bc)\ge 0◼\end{array}\end{array}$.

DAFTAR PUSTAKA

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