Inegalitate simpluta, 2

[Claudiu Mindrila]

Fie \( a, \ b, \ c>0 \) cu \( a^2+b^2+c^2=1 \). Sa se arate ca \( abc\left(a+b+c\right)+2\left(a^2b^2+b^2c^2+c^2a^2\right) \le 1 \).

Cristinel Mortici

[Mateescu Constantin]

Folosind succesiv inegalitatea \( x^2+y^2+z^2\ge xy+yz+zx \) obtinem :

\( a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\ge abc(a+b+c) \)

Asadar, \( LHS\le a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=(a^2+b^2+c^2)^2=1 \) .