Inequality 15

Problem: let  be positive reel numbers; Prove that:

\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\geq \frac{9}{4(a+b+c)}

Solution: assuming that,   we get ,

\frac{1}{(b+c)^2} \geq \frac{1}{(c+a)^2} \geq \frac{1}{(a+b)^2}    then by Chebychev,

S=\sum \frac{a}{(b+c)^2}\geq \frac{1}{3}(a+b+c)\left(\sum\frac{1}{(a+b)^2}\right)

and by Cauchy-Scwarz tow times,

\sum \frac{1}{(a+b)^2}\geq \frac{1}{3}\left(\frac{1}{a+b}\right)^2\geq \frac{1}{3}\left(\frac{9}{2(a+b+c)}\right)^2

then S\geq \frac{1}{3}(a+b+c)\left(\frac{1}{3}\left(\frac{9}{2(a+b+c)}\right)^2\right)=\frac{9}{4(a+b+c)}

Published in: on avril 7, 2009 at 7:10  Laisser un commentaire  

Inequality 14

Problem: let   be a positive reel numbers such as,   . Prove that;

ax+by+cz+2\sqrt{(xy+yz+zx)(ab+bc+ca)}\leq a+b+c

Ukraine, 2001

Solution: we can rewrite the inequality as:

2\sqrt{(xy+yz+zx)(ab+bc+ca)}\leq a(1-x)+b(1-y)+c(1-z)

or again ,2\sqrt{(xy+yz+zx)(ab+bc+ca)}\leq a(y+z)+b(z+x)+c(x+y)

assume that                and      ,

we get that    then by Chebychev,

a(x+y)+b(y+z)+c(z+x)\geq \frac{2}{3}(a+b+c)     (*)

and we have that,  xy+yz+zx\leq \frac{(x+y+z)^2}{3}=\frac{1}{3}

and  ab+bc+ca \leq \frac{(a+b+c)^2}{3}   then,

2\sqrt{(xy+yz+zx)(ab+bc+ca)}\leq \frac{2}{3}(a+b+c) 

from (*) and (**) we deduct that,

2\sqrt{(xy+yz+zx)(ab+bc+ca)} \leq a(x+y)+b(y+z)+c(z+x)

\Leftrightarrow ax+by+cz+2\sqrt{(xy+yz+zx)(ab+bc+ca)} \leq a+b+c

Published in: on avril 7, 2009 at 6:47  Laisser un commentaire  

Inequality 14

Problem: let   such as  , find the maximum of:                                            

Solution: first we remarque that,

T=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)

put that,   and    therefore

we have          2p\leq p^2+1=2+2q \Leftrightarrow p\leq q+1 

thus, p(1-q)\leq (1-q)(1+q)=1-q^2\leq 1 

then  equality holds when ,

Published in: on avril 7, 2009 at 2:51  Laisser un commentaire  

Inequality 13

Problem: if you know that the equation     has at least one reel root, prove that ,  

Tournament of the Towns, 1993

Solution: let  be a root of the equation, therefore, by Cauchy-Schwarz,

(a^2+b^2)(x^6+x^2)\geq (ax^3+bx)^2=(x^2+1)^4   and we have that,

           \Leftrightarrow x^8+6x^4+1\geq 4(x^6+x^2)

\Leftrightarrow x^8+4x^6+6x^4+4x^2+1 \geq 8(x^6+x^2)\Leftrightarrow (x^2+1)^4\geq 8(x^6+x^2)

then    a^2+b^2\geq \frac{(x^2+1)^4}{x^6+x^2}\geq8

Published in: on avril 7, 2009 at 2:25  Laisser un commentaire  

Inequality 12

Probelm: let  be  positive reel numbers such as, , Prove that:

\frac{b+c}{\sqrt{a}}+\frac{a+c}{\sqrt{b}}+\frac{b+a}{\sqrt{c}} \geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3

Gazeta matematicã

Solution:

assume that,     therefore by Chebychev,

S=\sum \frac{b+c}{\sqrt{a}} \geq \frac{2}{3}(a+b+c)\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)

Be Cauchy-Schwarz,    \frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\geq \frac{9}{\sqrt{a}+\sqrt{b}+\sqrt{c}}

and also   a+b+c\geq \frac{1}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2, then

S\geq \frac{2}{9}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\right)\left( \frac{9}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\right)=2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)

\Leftrightarrow \frac{1}{2}S\geq \sqrt{a}+\sqrt{b}+\sqrt{c}    (*)

By AM-GM,   S=\sum \frac{b+c}{\sqrt{a}} \geq 3\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{\sqrt{abc}}}

and  then S\geq 6 \Leftrightarrow \frac{1}{2}S \geq 3 (**)

After summation of (*) and (**) we get,

\frac{b+c}{\sqrt{a}}+\frac{a+c}{\sqrt{b}}+\frac{b+a}{\sqrt{c}} \geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3

Published in: on avril 7, 2009 at 1:35  Laisser un commentaire  

Inequality 11

Problem: let  , Prove that:

\sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)} < 1

Junior TST 2002,Romania

Solution: Put that, , therefore, by AM-GM

\sqrt{abc}\leq \left(\frac{x}{3}\right)^{\frac{3}{2}} and \sqrt{(1-a)(1-b)(1-c)}\leq \left(\frac{3-x}{3}\right)^{\frac{3}{2}}

and we have,         if       

then              \left(\frac{x}{3}\right)^{\frac{3}{2}}+\left(\frac{3-x}{3}\right)^{\frac{3}{2}} < \left(\frac{x+3-x}{3}\right)^{\frac{3}{2}}=1  

wich implies that, \sqrt{abc}+\sqrt{(1-a)(1-b)(1-c)}<1

Published in: on avril 7, 2009 at 1:11  Laisser un commentaire  

Inequality 10

Problem: let  , prove that:

\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \frac{3\sqrt{2}}{2}

Solution: Be Cauchy-Schwarz

\Leftrightarrow \sqrt{x^2+y^2} \geq \frac{x+y}{\sqrt{2}}   then; \sqrt{c^2+(1-a)^2}\geq \frac{1+c-a}{\sqrt{2}}

\sqrt{a^2+(1-b)^2}\geq \frac{1+a-b}{\sqrt{2}}    and   \sqrt{b^2+(1-c)^2}\geq \frac{1+b-c}{\sqrt{2}}

after summation we get;

\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq \frac{3}{\sqrt{2}}=\frac{3\sqrt{2}}{2}

Published in: on avril 7, 2009 at 1:00  Laisser un commentaire  

Problem 1

Problem: Find all positive integers  such as

   and  are all integers,

Solution:

\begin{cases}\frac{ab-1}{c}=k\\ \frac{bc-1}{a}=l\\ \frac{ca-1}{b}=m\end{cases}

Multiplying all equations we get \frac{a^2b^2c^2-a^2bc-ab^2c-abc^2+ab+bc+ca-1}{abc}=klm
\frac{ab+bc+ca-1}{abc}=klm+a+b+c-abc

\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc} is an integer.we have \frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{abc} = k where is an integer
suppose that one of is equal to for example then,
is integer then is integer wich implies (*)
is integer then is integer wich implies and from wich is false. therefore thus
k = \frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{abc}\leq \frac {3}{2} therefore or
if then \frac {1}{a} + \frac {1}{b} + \frac {1}{c} = \frac {1}{abc}
\Leftrightarrow ab + bc + ca = 1 wich is false, thus,
\frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{abc} = 1
\Leftrightarrow \frac {1}{bc}\left(\frac {bc - 1}{a} + b + c\right) = 1
and we have therefore,
1 = \frac {1}{bc}\left(\frac {bc - 1}{a} + b + c\right)\leq \frac {1}{bc}\left(\frac {bc - 1}{2} + b + c\right)
\Leftrightarrow bc \leq bc + 1 \leq 2(b + c) \Leftrightarrow \frac {1}{2} \leq \frac {1}{b} + \frac {1}{c}
suppose that therefore, \frac {1}{2} \leq \frac {1}{b} + \frac {1}{c}\leq \frac {2}{c} \Leftrightarrow 2\leq c\leq 4
if then, \frac {1}{a} + \frac {1}{b} - \frac {1}{4ab} = \frac {3}{4}
\Leftrightarrow (3b - 4)(3a - 4) = 13 wich don’t have any integer solution.
the or
if
\frac {1}{a} + \frac {1}{b} - \frac {1}{2ab} = \frac {1}{2}
\Leftrightarrow (b - 2)(a - 2) = 3
then
if
\frac {1}{a} + \frac {1}{b} - \frac {1}{3ab} = \frac {2}{3}
\Leftrightarrow (2b - 3)(2a - 3) = 7
then
finally the only solution for the problem is , with all permutation possible.

Published in: on avril 6, 2009 at 8:05  Laisser un commentaire  

inequality 9

Problem: let , such as  prove that:

4(ab+bc+ca) +\frac{a^2b^2}{a+b}+\frac{a^2c^2}{a+c}+\frac{b^2c^2}{b+c} \leq \frac{27}{2}

Solution:  let put and and
then the inequality becomes;
S=4\sum xy +3\sum \frac{x^2y^2}{x+y} \leq \frac{3}{2} with .
We have \frac{x^2y^2}{x+y} \leq \frac{xy(x+y)}{4} because
then S\leq 4\sum xy +\frac{3}{4}\left(\sum xy(x+y)\right) = 4\sum+\frac{3}{4}\left(\sum xy  -3xyz\right)
\Leftrightarrow S\leq \frac{19}{4}\left(\sum xy \right) - \frac{9xyz}{4}
By Shur, \sum xy \leq \frac{1+9xyz}{4} \Leftrightarrow \frac{19}{4}\left(\sum xy \right) \leq \frac{19+171xyz}{16}
then by AM-GM therefore:

S\leq \frac{19+\frac{135}{27}}{16}=\frac{3}{2}.

Published in: on mars 13, 2009 at 12:02  Laisser un commentaire  

Inequality 8

Published in: on mars 4, 2009 at 6:56  Laisser un commentaire  
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