Question

在三角形ABC中,求证：a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)>a^3+b^3+c^3+2ab

Answer 1

答：

a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)-(a^3+b^3+c^3)

=a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2-c^2)

=a*2bccosA+b*2accosB+c*2abcosC

=2abc(cosA+cosB+cosC）

=2abc[cosA+cosB+cos(180°-A-B)]

=2abc[cosA+cosB-cos(A+B)]

=2abc{2cos[(A+B)/2]*cos[(A-B)/2]-2[cos((A+B)/2)]^2+1}

=2abc{2cos[(A+B)/2]*[cos(A-B)/2-cos(A+B)/2]+1}

=2abc{2cos[(A+B)/2]*2sin(A/2)*sin(B/2)+1}

=2abc{4cos[(A+B)/2]*sin(A/2)*sin(B/2)+1}

因为：0°1

即：

a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)-(a^3+b^3+c^3)

=2abc{4cos[(A+B)/2]*sin(A/2)*sin(B/2)+1}

>2abc

所以：a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)>a^3+b^3+c^3+2abc