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)
[tex](x+y+z)(\frac{1}{x} +\frac{1}{y} +\frac{1}{z} )=[/tex]
[tex]=\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} =[/tex]
[tex]=1 +\frac{x}{y} + \frac{x}{z} +\frac{y}{x} +1 +\frac{y}{z} + \frac{z}{x} +\frac{z}{y} +1 =[/tex]
[tex]=3 +\frac{x}{y} +\frac{y}{x} + \frac{x}{z}+ \frac{z}{x} +\frac{y}{z} +\frac{z}{y} =[/tex]
[tex]=3 +\frac{x^2 + y^2}{xy} \frac{x^2+z^2}{xz}+\frac{y^2+z^2}{yz}=[/tex]
[tex]=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}=[/tex]
[tex]=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} =[/tex]
[tex]=3 +\frac{(x-y)^2}{xy}+2 +\frac{(x-z)^2}{xz}+2 +\frac{(y-z)^2}{yz}+2 =[/tex]
[tex]=9 +\frac{(x-y)^2}{xy} +\frac{(x-z)^2}{xz} +\frac{(y-z)^2}{yz} \geq 9[/tex]
(deoarece x, y, z sunt numere pozitive, iar la numaratorii fractiilor avem patrate, care sunt tot numere pozitive)
Asadar:
[tex](x+y+z)(\frac{1}{x} +\frac{1}{y} +\frac{1}{z} )\geq 9[/tex]
[tex](|a|+|b|+|c|)(\frac{1}{|a|} +\frac{1}{|b|} +\frac{1}{|c|} )\geq 9[/tex]
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:
[tex]A = \frac{x^2 + y^2 + z^2}{xyz} - \frac{1}{x} -\frac{1}{y} -\frac{1}{z} =[/tex]
[tex]= \frac{x^2 + y^2 + z^2}{xyz} - \frac{yz}{xyz} -\frac{xz}{xyz} -\frac{xy}{xyz} =[/tex]
[tex]= \frac{x^2 + y^2 + z^2 - yz - xz - xy}{xyz}=[/tex]
[tex]= \frac{2(x^2 + y^2 + z^2 - yz - xz - xy)}{2xyz}=[/tex]
[tex]= \frac{2x^2 + 2y^2 + 2z^2 - 2yz - 2xz - 2xy)}{2xyz}=[/tex]
[tex]= \frac{x^2 + x^2 +y^2 +y^2+ z^2 +z^2- 2yz - 2xz - 2xy}{2xyz}=[/tex]
[tex]= \frac{x^2 - 2xy+y^2 + x^2 - 2xz+ z^2+y^2 - 2yz+z^2 }{2xyz}=[/tex]
[tex]= \frac{(x-y)^2 + (x-z)^2+ (y-z)^2 }{2xyz} \geq 0[/tex]
(deoarece x, y, z sunt numere pozitive, iar la numaratorul fractiel avem sume de patrate, care sunt tot numere pozitive)
Asadar A > 0 , deci
[tex]\frac{x^2 + y^2 + z^2}{xyz} - \frac{1}{x} -\frac{1}{y} -\frac{1}{z} \geq 0[/tex]
[tex]\frac{x^2 + y^2 + z^2}{xyz} \geq \frac{1}{x} +\frac{1}{y} +\frac{1}{z}[/tex]
[tex]\frac{1}{x} +\frac{1}{y} +\frac{1}{z}\leq \frac{x^2 + y^2 + z^2}{xyz}[/tex]
[tex]\frac{1}{|a|} +\frac{1}{|b|} +\frac{1}{|c|}\leq \frac{a^2 + b^2 + c^2}{|abc|}[/tex]
