Jump to content
Sign in to follow this  
Louiscahyadi

Maraton Ketaksamaan

Recommended Posts

Soal ekuivalen dengan membuktikan $a^4+b^4+c^4 \geq a^2bc+ab^2c+abc^2$ yang jelas diperoleh dari ketaksamaan AM-GM seperti berikut: $a^4+a^4+b^4+c^4 \geq 4a^2bc$, $a^4+b^4+b^4+c^4 \geq 4ab^2c$, dan $a^4+b^4+c^4+c^4 \geq 4abc^2$.



Misalkan $a, b,$ dan $c$ adalah sisi-sisi dari segitiga $ABC$, buktikan bahwa $$ 4 \sqrt3 [ABC] \leq a^2+b^2+c^2-(a-b)^2-(b-c)^2-(c-a)^2$$.

Edited by Prihandoko
nambahin spoiler, merge post

Share this post


Link to post
Share on other sites

Dengan substitusi Ravi maka ketaksamaan pada soal ekuivalen dengan $$ xy + yz +zx \geq \sqrt{3xyz(x+y+z)}$$. sehingga cukup dibuktikan $$ x^{2}y^{2} + y^2 z^2 + z^2 x^2 \geq xyz(x+y+z)$$


Dengan AM-GM didapat
$$x^2 y^2 + y^2 z^2 \geq 2xy^{2}z$$ $$y^2 z^2 + z^2 x^2 \geq 2xyz^{2}$$ $$z^2 x^2 + x^2 y^2 \geq 2x^{2}yz$$
dengan menjumlahkan ketiganya maka terbukti




Buktikan bahwa $$ \frac{b}{a+2b+c} +\frac{c}{b+2c+d} +\frac{d}{c+2d+a} +\frac{a}{d+2a+b} \leq 1$$ untuk $a,b,c,d \geq 0$

Edited by Prihandoko
merge post

Share this post


Link to post
Share on other sites

$2 \geq 2LHS$

$4 \geq 2LHS+2$

$4-2LHS \geq 2$

$(a+c)(\frac{1}{a+2b+c}+\frac{1}{a+2d+c})+(b+d)(\frac{1}{b+2a+d}+\frac{1}{b+2c+a}) \geq 2$

CS

$(a+c+b+d)(\frac{4}{2a+2b+2c+2d}) \geq 2$

Jelas benar

Buktikan $2[5,0,0]+2[3,2,0]+[3,1,1] \geq 5[4,1,0]$

Edited by jci_133

Share this post


Link to post
Share on other sites

Dari Schur,kita punya $[5,0,0]+[3,1,1]\geq 2[4,1,0]$.Selanjutnya,perhatikan bahwa:$$[5,0,0]-3[4,1,0]+2[3,2,0]=\frac{1}{12}\sum_{sym}x^{5}-3x^{4}y+2x^{3}y^{2}+2x^{2}y^{3}-3xy^{4}+y^{5}=\frac{1}{12}\sum_{sym}(x-y)^{4}(x+y)\geq 0$$ tinggal menjumlahkan keduanya

perbedaan ke 5:

$a,b,c$ riil positif dimana $a+b\geq c,b+c\geq a,c+a\geq b$.Buktikan bahwa:$$2a^{2}(b+c)+2b^{2}(c+a)+2c^{2}(a+b)\geq a^3+b^3+c^3+9abc$$

Edited by Mary K

Share this post


Link to post
Share on other sites

Wlog $a \leq b \leq c$

Case 1: $a+b=c$

Ganti semua $c$ dalam $a+b$

Setelah bongkar akhirnya $a^3 + b^3 \geq a^2b + ab^2$ AM-GM cukup

Case 2: $a+b > c$

Ravii

$a=x+y , b=y+z, c=z+x$

BONGKAR!!!

Ujung"nya $[3,0,0]+[1,1,1] \geq 2[2,1,0]$ scuur itu

a,b,c sisi-sisi segitiga r jari-jari lingkaran dalam

Buktikan $\frac{1}{r} \geq \frac{1}{a} + \frac{1}{b}+\frac{1}{c}$

Edited by jci_133

Share this post


Link to post
Share on other sites

Definisikan $A$ sebagai luas dan $h_{x}$ sebagai garis tinggi dari titik $X$.Maka:$$\sum h_{a}\leq \frac{1}{2}\sum b+c\Rightarrow \sum\frac{A}{a}\leq \frac{1}{2}\sum a\Rightarrow \sum\frac{1}{a}\leq \frac{1}{r}$$

7.Buktikan untuk sembarang riil positif $a,b,c$ berlaku:$$a^3+b^3+c^3+3abc\geq \sum_{cyc}ab\sqrt{2(a^2+b^2)}$$

Edited by Mary K

Share this post


Link to post
Share on other sites

Misalkan $a,b,c$ bilangan real positiv sehingga $a^2+b^2+c^2=3$.

Buktikan

$ \sqrt[4]{5a^2+4(b+c)+3} +\sqrt[4]{5b^2+4(c+a)+3} +\sqrt[4]{5c^2+4(a+b)+3} \leq 6 $

Share this post


Link to post
Share on other sites

Dari Pertidaksamaan Power Mean dan QM-AM,didapat:$$\sum_{cyc}\sqrt[4]{5a^2+4(b+c)+3}\leq 3\sqrt[4]{\frac{\sum_{cyc}5a^2+4(b+c)+3}{3}}\leq 6$$

Diketahui $\frac{1}{3}\leq x,y,z\leq 3$. Buktikan bahwa $$\sum_{cyc}\frac{x}{x+y}\geq\frac{7}{5}$$

Edited by Mary K

Share this post


Link to post
Share on other sites

Wlog x min, y,z aku fixed.

Aku bagi kasus:

*$y\geq z\geq x$

Maka

$f(x,y,z)=\frac{x}{x+y}+\frac{z}{x+z}+c$

$\geq\frac{z-y}{x+(yz/x)+y+z}+c+1$

utk dpt nilai $\frac{z-y}{x+(yz/x)+y+z}+c+1$ min maka $x+yz/x$ hrs min.

pake AMGM, $\frac{z-y}{x+(yz/x)+y+z}+c+1\geq\frac{z-y}{2\sqrt{yz}+y+z}+[y/(y+z)+1]$

$\geq(\frac{1}{1+z/y}+\frac{2}{\sqrt{y}{z}+1})$

mis $ y/z=k^2 \leq(9)$ implies $k\leq3$

Prove $1/(\frac{1}{k}+1)+2/(k+1)\geq7/5$

tp kesamaan ny jls g terbukti krn nti didpt y,z,x bruturut2 3,1/3,1. X bkn min

*utk $z\geq y\geq x$

$f(x,y,z)=\frac{x}{x+y}+\frac{z}{x+z}+c$

$\geq\frac{z-y}{x+(yz/x)+y+z}+c+1$

utk dpt nilai $\frac{z-y}{x+(yz/x)+y+z}+c+1$ min maka $x+yz/x$ hrs max y trjd saat x=1/3.

Bandingkan utk y,x fix maka f(x,y,z) min saat $z+xy/z$ max yaitu saat z=3.

kasus f(x,y,z)min saat z=3,x=1/3, prove itu msh lbh bsr dr 7/5 dg keaamaan saat y=1.

Buktikan utk a,b,c real positif,

$\sum_{cyc} \frac{a}{\sqrt{a+b}} \geq \frac{1}{\sqrt{2}} (\sum\sqrt{a})$

Edited by Prihandoko

Share this post


Link to post
Share on other sites

Wlog x min, y,z aku fixed.

Aku bagi kasus:

*$y\geq z\geq x$

Maka

$f(x,y,z)=\frac{x}{x+y}+\frac{z}{x+z}+c$

$\geq\frac{z-y}{x+(yz/x)+y+z}+c+1$

utk dpt nilai $\frac{z-y}{x+(yz/x)+y+z}+c+1$ min maka $x+yz/x$ hrs min.

pake AMGM, $\frac{z-y}{x+(yz/x)+y+z}+c+1\geq\frac{z-y}{2\sqrt{yz}+y+z}+[y/(y+z)+1]$

$\geq(\frac{1}{1+z/y}+\frac{2}{\sqrt{y}{z}+1})$

mis $ y/z=k^2 \leq(9)$ implies $k\leq3$

Prove $1/(\frac{1}{k}+1)+2/(k+1)\geq7/5$

tp kesamaan ny jls g terbukti krn nti didpt y,z,x bruturut2 3,1/3,1. X bkn min

*utk $z\geq y\geq x$

$f(x,y,z)=\frac{x}{x+y}+\frac{z}{x+z}+c$

$\geq\frac{z-y}{x+(yz/x)+y+z}+c+1$

utk dpt nilai $\frac{z-y}{x+(yz/x)+y+z}+c+1$ min maka $x+yz/x$ hrs max y trjd saat x=1/3.

Bandingkan utk y,x fix maka f(x,y,z) min saat $z+xy/z$ max yaitu saat z=3.

kasus f(x,y,z)min saat z=3,x=1/3, prove itu msh lbh bsr dr 7/5 dg keaamaan saat y=1.

Buktikan utk a,b,c real positif,

$\sum_{cyc} \frac{a}{\sqrt{a+b}} \geq \frac{1}{\sqrt{2}} (\sum\sqrt{a})$

 

ini soal repost tolong ganti soalnya 

http://olimpiade.org/forum/index.php/topic/316-terlihat-mudah-padahal/

  • Upvote 1

Share this post


Link to post
Share on other sites

Jika $a,b$ positif real, buktikan bahwa $$2\sqrt{a} + 3 \sqrt[3]{b} \ge 5\sqrt[5]{ab}$$

AM-GM:$$\frac{\sqrt{a}+\sqrt{a}+\sqrt[3]{b}+\sqrt[3]{b}+\sqrt[3]{b}}{5}\geq \sqrt[5]{ab}\Rightarrow 2\sqrt{a}+3\sqrt[3]{b}\geq 5\sqrt[5]{ab}$$. Soal baru: Jika $x,y\in{R}$, Tentukan minimum dari $$\sum_{i=1}^{10}\sum_{j=1}^{10}\sum_{k=1}^{10}|k(x+y-10i)(3x-6y-36j)(19x+95y-95k)|$$ Edited by SENA

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this  

×