Question

A7. Puteti sa ma ajutați va rog la ex. Ăsta ?

Answer 1

Răspuns:

Explicație pas cu pas:

Ca sa fie mai usor de scris, hai sa scapam de module

asadar, notam |a| = x  ; |b| = y  ; |c| = z, cu x,y,z numere reale pozitive, diferite de 0

a)

$(x+y+z)(\frac{1}{x} +\frac{1}{y} +\frac{1}{z} )=$

$=\frac{x}{x} +\frac{x}{y} +\frac{x}{z} +\frac{y}{x} +\frac{y}{y} +\frac{y}{z} + \frac{z}{x} +\frac{z}{y} +\frac{z}{z} =$

$=1 +\frac{x}{y} + \frac{x}{z} +\frac{y}{x} +1 +\frac{y}{z} + \frac{z}{x} +\frac{z}{y} +1 =$

$=3 +\frac{x}{y} +\frac{y}{x} + \frac{x}{z}+ \frac{z}{x} +\frac{y}{z} +\frac{z}{y} =$

$=3 +\frac{x^2 + y^2}{xy} \frac{x^2+z^2}{xz}+\frac{y^2+z^2}{yz}=$

$=3 +\frac{x^2 + y^2-2xy+2xy}{xy} +\frac{x^2+z^2-2xz+2xz}{xz}+\frac{y^2+z^2-2yz+2yz}{yz}=$

$=3 +\frac{x^2 + y^2-2xy}{xy}+\frac{2xy}{xy} +\frac{x^2+z^2-2xz}{xz}+\frac{2xz}{xz} +\frac{y^2+z^2-2yz}{yz}+\frac{2yz}{yz} =$

$=3 +\frac{(x-y)^2}{xy}+2 +\frac{(x-z)^2}{xz}+2 +\frac{(y-z)^2}{yz}+2 =$

$=9 +\frac{(x-y)^2}{xy} +\frac{(x-z)^2}{xz} +\frac{(y-z)^2}{yz} \geq 9$

(deoarece x, y, z sunt numere pozitive, iar la numaratorii fractiilor avem patrate, care sunt tot numere pozitive)

Asadar:

$(x+y+z)(\frac{1}{x} +\frac{1}{y} +\frac{1}{z} )\geq 9$

$(|a|+|b|+|c|)(\frac{1}{|a|} +\frac{1}{|b|} +\frac{1}{|c|} )\geq 9$

b)

observam ca a² = |a|² = x², respectiv  b² = |b|² = y² si  c² = |c|² = z²

si

|abc| = |a|*|b|*|c| = x*y*z

Notam cu A expresia:

$A = \frac{x^2 + y^2 + z^2}{xyz} - \frac{1}{x} -\frac{1}{y} -\frac{1}{z} =$

$= \frac{x^2 + y^2 + z^2}{xyz} - \frac{yz}{xyz} -\frac{xz}{xyz} -\frac{xy}{xyz} =$

$= \frac{x^2 + y^2 + z^2 - yz - xz - xy}{xyz}=$

$= \frac{2(x^2 + y^2 + z^2 - yz - xz - xy)}{2xyz}=$

$= \frac{2x^2 + 2y^2 + 2z^2 - 2yz - 2xz - 2xy)}{2xyz}=$

$= \frac{x^2 + x^2 +y^2 +y^2+ z^2 +z^2- 2yz - 2xz - 2xy}{2xyz}=$

$= \frac{x^2 - 2xy+y^2 + x^2 - 2xz+ z^2+y^2 - 2yz+z^2 }{2xyz}=$

$= \frac{(x-y)^2 + (x-z)^2+ (y-z)^2 }{2xyz} \geq 0$

(deoarece x, y, z sunt numere pozitive, iar la numaratorul fractiel avem sume de patrate, care sunt tot numere pozitive)

Asadar A > 0 , deci

$\frac{x^2 + y^2 + z^2}{xyz} - \frac{1}{x} -\frac{1}{y} -\frac{1}{z} \geq 0$

$\frac{x^2 + y^2 + z^2}{xyz} \geq \frac{1}{x} +\frac{1}{y} +\frac{1}{z}$

$\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\leq \frac{x^2 + y^2 + z^2}{xyz}$

$\frac{1}{|a|} +\frac{1}{|b|} +\frac{1}{|c|}\leq \frac{a^2 + b^2 + c^2}{|abc|}$