Question

Dacă $a,b,c \geq 0,$ arătați că $(a+b+c)^{4} \geq 16( a^{2} b^{2} + b^{2} c ^{2} + c^{2} a^{2} ).$

Answer 1

[tex](a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)(1) \\ Folosind\ inegalitatea\ dintre\ media\ geometrica\ si\ media\ aritmetica:\\ x+y \geq 2 \sqrt{xy} \\ Din\ (1)=>(a+b+c)^2 \geq 2 \sqrt{(a^2+b^2+c^2)\cdot 2(ab+ac+bc)} \\ Inegalitatea\ anterioara\ o \ ridicam\ la\ patrat:\\ (a+b+c)^4 \geq 8 (a^2+b^2+c^2)\cdot (ab+ac+bc)(2)\\ (a^2+b^2+c^2)\cdot (ab+ac+bc)=ab(a^2+b^2)+ac(a^2+c^2)+bc(a^2+c^2)\\ +abc^2+acb^2+bca^2 \geq2a^2b^2+2a^2c^2+2b^2c^2(3) \\ Din\ (2)+(3)=>(a+b+c)^4 \geq 16(a^2b^2+a^2c^2+b^2c^2) [/tex]