Maths Olympiad Prep

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Algebra Difficulty 5.6 AIME, harder Prove it

Example 5. In △ABC\triangle ABC, a,b,ca, b, c are the lengths of its three sides. Prove that:
a2b(a−b)+b2c(b−c)+c2a(c−a)⩾0. \begin{array}{l} a^{2} b(a-b)+b^{2} c(b-c) \\ +c^{2} a(c-a) \geqslant 0 . \end{array}

and specify when equality holds. (24th IMO problem)

Solution

 Proof  Without loss of generality, assume a⩾b⩾c>0, Let x=a+b−c2,y=a+c−b2,z=a+a+b−c2, Then a=y+z,b=x+z,c=x+y, If x=0,y>0,c>0.a2b(c−b)+b2c(b−c)+c2a(c−a)=(y+z)2(x+z)(y−x)+(x+z)2⋅(x+y)(z−y)+(x+y)2(y+z)(x−z)=2yz(x2+z2)+2xz(x2+y2)+2xy(y2+z2)−4xy2z−4xyz2−4x2yz⩾4xyz2+4x2yz+4xy2z−4x2yz−4xyz2−4x2yz=0.\begin{array}{c}\text { Proof } \text { Without loss of generality, assume } a \geqslant b \geqslant c>0, \\ \text { Let } x=\frac{a+b-c}{2}, y=\frac{a+c-b}{2}, z=a+\frac{a+b-c}{2}, \\ \text { Then } \quad a=y+z, b=x+z, c=x+y, \\ \text { If } x=0, y>0, c>0 . \\ a^{2} b(c-b)+b^{2} c(b-c)+c^{2} a(c-a) \\ =(y+z)^{2}(x+z)(y-x)+(x+z)^{2} \\ \cdot(x+y)(z-y)+(x+y)^{2}(y+z)(x-z) \\ =2 y z\left(x^{2}+z^{2}\right)+2 x z\left(x^{2}+y^{2}\right) \\ +2 x y\left(y^{2}+z^{2}\right)-4 x y^{2} z-4 x y z^{2}-4 x^{2} y z \\ \geqslant 4 x y z^{2}+4 x^{2} y z+4 x y^{2} z-4 x^{2} y z \\ \quad-4 x y z^{2}-4 x^{2} y z=0 .\end{array}
 When and only when x2=y2,y2=z2,z2=x2,  i.e., x=y=z, or equivalently a=b=c, the equality holds. \begin{array}{l}\text { When and only when } x^{2}=y^{2}, y^{2}=z^{2}, z^{2}=x^{2} \text {, } \\ \text { i.e., } x=y=z \text {, or equivalently } a=b=c \text {, the equality holds. }\end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.