Mathsmaster’s Blog

Inequality 15

Problem: let $a,b,c$ 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, $a \geq b \geq c$ 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 $S\geq \frac{1}{3}(a+b+c)\left(\frac{1}{3}\left(\frac{9}{2(a+b+c)}\right)^2\right)=\frac{9}{4(a+b+c)}$

Publié dans :

Inégalités

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

Inequality 14

Problem: let $a,b,c,x,y,z$ be a positive reel numbers such as, $x+y+z=1$ . 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 $a\geq b \geq c$ and $x \geq y \geq z$ ,

we get that $x+y \geq x+z \geq y+z$ 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,

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

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

Publié dans :

Inégalités

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

Inequality 14

Problem: let $x,y,z \in\mathbb{R}$ such as $x^2+y^2+z^2=1$ , find the maximum of: $T=x^3+y^3+z^3-3xyz$

Solution: first we remarque that,

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

put that, $p=x+y+z$ and $q=xy+yz+zx$ therefore $T=p(1-q)$

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 $T_{max}=1$ equality holds when $x=1$ , $y=z=0$

Publié dans :

Inégalités

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

Inequality 13

Problem: if you know that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one reel root, prove that , $a^2+b^2\geq 8$

Tournament of the Towns, 1993

Solution: let $x$ 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,

$(x^2 -1)^4\geq 0$ $\Leftrightarrow x^8+6x^4+1\geq 4(x^6+x^2)$

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

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

Publié dans :

Inégalités

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

Inequality 12

Probelm: let $a,b,c$ be positive reel numbers such as, $abc=1$ , 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, $a \geq b \geq c$ 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 $(a+b)(b+c)(c+a)\geq 8abc=8$ then $S\geq 6 \Leftrightarrow \frac{1}{2}S \geq 3$ (**)

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

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

Publié dans :

Inégalités

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

Inequality 11

Problem: let $a,b,c \in [0,1]$ , Prove that:

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

Junior TST 2002,Romania

Solution: Put that, $x=a+b+c$ , 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, $n^a+m^a< (n+m)^a$ if $a>1$

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

Publié dans :

Inégalités

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

Inequality 10

Problem: let $a,b,c \in \mathbb{R}$ , 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 $2(x^2+y^2)\geq (x+y)^2$

$\Leftrightarrow \sqrt{x^2+y^2} \geq \frac{x+y}{\sqrt{2}}$ 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;

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

Publié dans :

Inégalités

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

Problem 1

Problem: Find all positive integers $(a,b,c)$ such as

$\frac{ab-1}{c},\frac{bc-1}{a}$ and $\frac{ca-1}{b}$ 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 $k$ is an integer
suppose that one of $a,b,c$ is equal to $1$ for example $c$ then,
$\frac {ac - 1}{b}$ is integer then $\frac {a - 1}{b}$ is integer wich implies $b\leq a - 1$ (*)
$\frac {bc - 1}{a}$ is integer then $\frac {b - 1}{a}$ is integer wich implies $a\leq b - 1$ and from $a\leq b - 1 \leq a - 2$ wich is false. therefore $a,b,c \geq 2$ thus
$k = \frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{abc}\leq \frac {3}{2}$ therefore $k = 1$ or $k = 0$
if $k = 0$ then $\frac {1}{a} + \frac {1}{b} + \frac {1}{c} = \frac {1}{abc}$
$\Leftrightarrow ab + bc + ca = 1$ wich is false, thus, $k = 1$
$\frac {1}{a} + \frac {1}{b} + \frac {1}{c} - \frac {1}{abc} = 1$
$\Leftrightarrow \frac {1}{bc}\left(\frac {bc - 1}{a} + b + c\right) = 1$
and we have $a\geq 2$ 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 $b\geq c$ therefore, $\frac {1}{2} \leq \frac {1}{b} + \frac {1}{c}\leq \frac {2}{c} \Leftrightarrow 2\leq c\leq 4$
if $c = 4$ 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 $c = 2$ or $c = 3$
if $c = 2$
$\frac {1}{a} + \frac {1}{b} - \frac {1}{2ab} = \frac {1}{2}$
$\Leftrightarrow (b - 2)(a - 2) = 3$
then $S = \{2,3,5\}$
if $c = 3$
$\frac {1}{a} + \frac {1}{b} - \frac {1}{3ab} = \frac {2}{3}$
$\Leftrightarrow (2b - 3)(2a - 3) = 7$
then $S = \{2,3,5\}$
finally the only solution for the problem is , $S = \{2,3,5\}$ with all permutation possible.

Publié dans :

Algèbre

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

inequality 9

Problem: let $a,b,c \in \mathbb{R^{+*}}$ , such as $a+b+c=3$ 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 $a=3x$ and $b=3y$ and $c=3z$
then the inequality becomes;
$S=4\sum xy +3\sum \frac{x^2y^2}{x+y} \leq \frac{3}{2}$ with $x+y+z=1$ .
We have $\frac{x^2y^2}{x+y} \leq \frac{xy(x+y)}{4}$ because $(x+y)^2 \geq 4xy$
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 $S\leq \frac{19+135xyz}{16}$ by AM-GM $xyz \leq \frac{1}{27}$ therefore: