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AdolfKevlar
@AdolfKevlar
July 2022
1
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Докажите, что для любых действительно чисел a,b справедливо неравенство:
[tex] {a}^{2} + {b}^{2} + {c}^{2} \geqslant ab + bc + ac[/tex]
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sangers
A²+b²+c²≥ab+bc+ac |×2
2a²+2b²+2c²≥2ab+2bc+2ac
a²+a²+b²+b²+c²+c²-2ab-2bc-2ac≥0
(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ac+c²)≥0
(a-b)²+(b-c)²+(a-c)²≡≥0.
1 votes
Thanks 1
AdolfKevlar
Большое Большое спасибо! !
sangers
Удачи.
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Answers & Comments
2a²+2b²+2c²≥2ab+2bc+2ac
a²+a²+b²+b²+c²+c²-2ab-2bc-2ac≥0
(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ac+c²)≥0
(a-b)²+(b-c)²+(a-c)²≡≥0.