Contoh Soal 15 (Segitiga dan Ketaksamaan)

$\begin{array}{l}\\ 71.& (\mathbf{\text{IMO 2001}})\\ & \text{Diberikan}a,b\text{dan}c\text{bilangan real positif, tunjukkan}\\ & {\displaystyle \frac{a}{\sqrt{a^2+8bc}}+\frac{b}{\sqrt{b^2+8ca}}+\frac{c}{\sqrt{c^2+8ab}}\ge 1}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\displaystyle \frac{1}{3}\left({\displaystyle \sum _{i=1}^3{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{1}{3}}\ge ({\displaystyle \sum _{\text{siklik}}^{.}a)}\\ & \text{dapat juga dituliskan lebih sederhana}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}^{\frac{1}{3}}{\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^{\frac{2}{3}}\ge \left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)\\ & \iff \left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right){\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3\\ & \iff {\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\displaystyle \frac{{\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3}{\left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)}}\\ & \text{Perhatikan bahwa}\\ & \bullet {\left({\displaystyle \sum _{\text{siklik}}^{.}a}\right)}^3=(a+b+c{)}^3\ge a^3+b^3+c^3+24abc\\ & \bullet \left({\displaystyle \sum _{\text{siklik}}^{.}a(a^2+8bc)}\right)=a^3+b^3+c^3+24abc\\ & \text{maka}\\ & {\left({\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}}}\right)}^2\ge {\displaystyle \frac{a^3+b^3+c^3+24abc}{a^3+b^3+c^3+24abc}=1}\\ & {\displaystyle \sum _{\text{siklik}}^{.}{\displaystyle \frac{a}{\sqrt{a^2+8bc}}\ge 1◼}}\end{array}\end{array}$.

$.\begin{array}{rl}& \mathbf{\text{Sebagai catatan}}\\ & (a+b+c{)}^3=a^3+b^3+c^3+3a^{2}b+3a{b}^2\\ & +3b^{2}c+3b{c}^2+3a^{2}c+3a{c}^2+6abc\\ & (a+b+c{)}^3-(a^3+b^3+c^3)=3a^{2}b+3a{b}^2\\ & +3b^{2}c+3b{c}^2+3a^{2}c+3a{c}^2+6abc\\ & \text{Dengan AM-GM akan diperoleh bentuk}\\ & (a+b+c{)}^3-(a^3+b^3+c^3)\\ & \ge 3(3\sqrt[3]{(abc{)}^3}+3\sqrt[3]{(abc{)}^3})+6abc\\ & \ge 3(3abc+3abc)+6abc=24abc\\ & \text{Sehingga}\\ & (a+b+c{)}^3\ge a^3+b^3+c^3+24abc\end{array}$.

$\begin{array}{l}\\ 72.& \text{Jika}a,b,\text{dan}c\text{bilangan real positif}\\ & \text{dengan}a+b+c=1\text{tunjukkan bahwa}\\ & {\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}\ge {\displaystyle \frac{81}{4}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}& \text{Dengan Ketaksamaan Holder pilih}\\ & {\lambda }_1={\lambda }_2={\lambda }_3={\lambda }_4={\displaystyle \frac{1}{4}\left({\displaystyle \sum _{i=1}^4{\lambda }_i=1}\right)\text{akan diperoleh}}\\ & (a+b+c)(b+c+a+c+a+b)(b+c+a+c+a+b)\left({\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}}\right)\\ & \ge (1^{{.}^{\frac{1}{4}}}+1^{{.}^{\frac{1}{4}}}+1^{{.}^{\frac{1}{4}}}{)}^4=3^4=81\\ & \iff (1)(2)(2)\left({\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}}\right)\ge 81\\ & \iff {\displaystyle \frac{1}{a(b+c{)}^2}+\frac{1}{b(a+c{)}^2}+\frac{1}{c(a+b{)}^2}\ge {\displaystyle \frac{81}{4}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 73.& \text{Buktikan bahwa setiap bilangan real}\\ & \text{positif}a,b\text{dan}c\text{berlaku}\\ & a^2+b^2+c^2\ge ab+ac+bc\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}& \text{Alternatif 1}\\ & \text{Perhatikan bahwa}(a-b{)}^2\ge 0\\ & (a-c{)}^2\ge 0,\text{dan}(b-c{)}^2\ge 0\\ & \text{adalah benar, maka}\\ & (a-b{)}^2=a^2-2ab+b^2\ge 0\\ & \iff a^2+b^2\ge 2ab.....(1)\\ & \text{Dengan cara yang kurang lebih sama}\\ & \text{akan didapatkan}\\ & \bullet a^2+c^2\ge 2ac.....(2)\\ & \bullet b^2+c^2\ge 2bc.....(1)\\ & \text{Jika ketaksamaan}(1),(2),\mathrm{\&}(3)\text{dijumlahkan}\\ & \text{akan didapatkan bentuk}\\ & 2a^2+2b^2+2c^2\ge 2ab+2ac+2bc\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 2}\\ & \text{Dengan ketaksamaan}\mathbf{\text{Cauchy-Schwarz}}\\ & (a^2+b^2+c^2)(b^2+c^2+a^2)\ge (ab+bc+ca{)}^2\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 3}\\ & \text{Untuk barisan}(a,b,c),\text{asumsikan}a\ge b\ge c\\ & \text{maka dengan}\mathbf{\text{Ketaksamaan Renata}}\text{diperoleh}\\ & a.a+b.b+c.c\ge ab+bc+ca\\ & a^2+b^2+c^2\ge ab+ac+bc◼\\ & \text{Alternatif 4}\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=0,\\ & \text{yaitu}\\ & a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b)\ge 0\\ & \iff a^0(a-b)(a-c)+b^0(b-a)(b-c)+c^0(c-a)(c-b)\ge 0\\ & \iff (a-b)(a-c)+(b-a)(b-c)+(c-a)(c-b)\ge 0\\ & \iff a^2+b^2+c^2-ab-ac-bc\ge 0\\ & \iff a^2+b^2+c^2\ge ab+ac+bc◼\end{array}\end{array}$.

$\begin{array}{l}\\ 74.& \mathbf{\text{(IMO 1964)}}\\ & \text{Jika}a,b,c\text{adalah panjang sisi-sisi segitiga}\\ & \text{tunjukkan bahwa}\\ & a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\le 3abc\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Dengan}\mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1.\\ & a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ca(c+a)\\ & \iff 3abc\ge ab(a+b)-a^3+bc(b+c)-b^3+ca(c+a)-c^3\\ & \iff 3abc\ge a^{2}b+a{b}^2-c^3+b^{2}c+b{c}^2-b^3+c^{2}a+c{a}^2-c^3\\ & \iff 3abc\ge a^{2}b+c{a}^2-a^3+a{b}^2+b^{2}c-b^3+c^{2}a+b{c}^2-c^3\\ & \iff 3abc\ge a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\\ & \text{atau}\\ & \iff a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)\le 3abc◼\end{array}$ .

$\begin{array}{l}\\ 75.& \mathbf{\text{(IMO 2000)}}\\ & \text{Jika}a,b,c\text{bilangan real positif dengan}\\ & abc=1,\text{tunjukkan bahwa}\\ & (a+1-{\displaystyle \frac{1}{b}})(b+1-{\displaystyle \frac{1}{c}})(c+1-{\displaystyle \frac{1}{a}})\le 1\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Misalkan}a={\displaystyle \frac{x}{y},b=\frac{y}{z},\text{dan}c={\displaystyle \frac{z}{x},\text{maka}}}\\ & \left({\displaystyle \frac{x}{y}+1-\frac{z}{y}}\right)\left({\displaystyle \frac{y}{z}+1-\frac{x}{z}}\right)\left({\displaystyle \frac{z}{x}+1-\frac{y}{x}}\right)\le 1\\ & \iff \left({\displaystyle \frac{x+y-z}{y}}\right)\left({\displaystyle \frac{y+z-x}{z}}\right)\left({\displaystyle \frac{z+x-y}{x}}\right)\le 1\\ & \iff (x+y-z)(y+z-x)(z+x-y)\le xyz,\text{atau}\\ & \iff xyz\ge (x+y-z)(y+z-x)(z+x-y)\\ & \text{Bentuk terakhir memenuhi bentuk kedua dari}\\ & \mathbf{\text{Ketaksamaan Schur}}\text{saat}r=1.\\ & \text{Jadi},\\ & (a+1-{\displaystyle \frac{1}{b}})(b+1-{\displaystyle \frac{1}{c}})(c+1-{\displaystyle \frac{1}{a}})\le 1◼\end{array}$.