Question

Mă ajutați, vă rog, cu problema 26 de mai jos? 

 

Mulțumesc.

Semnul este ≥.

Answer 1

$(a+b+c)*( \frac{1}{a} + \frac{1}{b} +\frac{1}{c} ) \geq 9$ ⇔
$(a+b+c)*(ab+ac+bc) \geq 9abc$⇔
$a^{2} b+ a^{2} c+abc+ b^{2} a+abc+ b^{2} c+abc+ c^{2} a+ c^{2} b \geq 9abc$⇔
$a^{2} b+ a^{2} c+ b^{2} a+ b^{2} c+ c^{2} a+ c^{2} b \geq6abc$⇔
$a^{2} b-abc+ a^{2} c-abc+ b^{2} a-abc+ b^{2} c-abc+ c^{2} a-abc+ c^{2} b-abc$$\geq 0$⇔$ab(a-c)+ac(a-b)+ab(b-c)+bc(b-a)+ac(c-b)+bc(c-a) \geq 0$⇔
$(a-c)(ab-bc)+(a-b)(ac-bc)+(b-c)(ab-ac) \geq 0$⇔$(a-c)^{2} b+(a-b)^{2} c+(b-c)^{2} a \geq 0$ Adevarat pt ca a, b, c pozitive si $(a-b)^{2} \geq 0,$ $(a-c)^{2} \geq 0,$ $(b-c)^{2} \geq 0.$