Question

Cum demonstrez :
√a+√b+√c >= radical din a+b+c ?

Answer 1

[tex]\text{Din propriet\u a\c tile radicalului:}\\ \sqrt{a+b}\leq\sqrt a+\sqrt b\\ \text{Deci,}\\ \sqrt{a+b+c}=\sqrt{a+(b+c)}\\ \text{Aplic\u am proprietatea:}\\ \sqrt{a+(b+c)}\leq\sqrt a+\sqrt{b+c}\\ \text{Aplic\^ and din nou proprietatea, avem c\u a}\\ \sqrt a+\sqrt{b+c}\leq\sqrt a + \sqrt b + \sqrt c\\ \text{Cum } \sqrt{a+b+c}\leq\sqrt a+\sqrt{b+c} \text{ \c si}\\ \sqrt a+\sqrt{b+c}\leq\sqrt a + \sqrt b + \sqrt c\\ \text{Rezult\u a c\u a:}\\ \sqrt{a+b+c}\leq\sqrt a + \sqrt b + \sqrt c[/tex]