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Algebra Difficulty 6.0 AIME, harder Prove it

Prove: For any positive real numbers a,b,c,da, b, c, d, we have
(a−b)(a−c)a+b+c+(b−c)(b−d)b+c+d+(c−d)(c−a)c+d+a+(d−a)(d−b)d+a+b⩾0, \begin{array}{l} \frac{(a-b)(a-c)}{a+b+c}+\frac{(b-c)(b-d)}{b+c+d}+ \\ \frac{(c-d)(c-a)}{c+d+a}+\frac{(d-a)(d-b)}{d+a+b} \geqslant 0, \end{array}

and determine the conditions under which equality holds.

Solution

Proof: Let x1=a+b+c,x2=b+c+dx_{1}=a+b+c, x_{2}=b+c+d,
x3=c+d+a,x4=d+a+b.  x_{3}=c+d+a, x_{4}=d+a+b \text {. }

Obviously, xi>0x_{i}>0, and
∣xi−xjxk∣<1(i≠j≠k∈{1,2,3,4}). \left|\frac{x_{i}-x_{j}}{x_{k}}\right|<1(i \neq j \neq k \in\{1,2,3,4\}) .

Then equation (1) is equivalent to
(x2−x3)(x2−x4)x1+(x3−x4)(x3−x1)x2+(x4−x1)(x4−x2)x3+(x1−x2)(x1−x3)x4⩾0. \begin{array}{l} \frac{\left(x_{2}-x_{3}\right)\left(x_{2}-x_{4}\right)}{x_{1}}+\frac{\left(x_{3}-x_{4}\right)\left(x_{3}-x_{1}\right)}{x_{2}}+ \\ \frac{\left(x_{4}-x_{1}\right)\left(x_{4}-x_{2}\right)}{x_{3}}+\frac{\left(x_{1}-x_{2}\right)\left(x_{1}-x_{3}\right)}{x_{4}} \geqslant 0 . \end{array}

By cyclic symmetry, without loss of generality, assume x4x_{4} is the largest. Thus,
the left side of the above equation
=(x1−x3)[(x3−x2)+(x1−x3)]x4−(x2−x3)(x4−x2)x1+(x3−x1)[(x2−x4)+(x3−x2)]x2+(x4−x2)[(x4−x2)+(x2−x3)+(x3−x1)]x3=(x3−x1)2x4+(x4−x2)2x3+(x3−x1)(x4−x2)(x3−x2)x2x4−(x3−x1)(x4−x2)(x3−x2)x2x3+(x3−x1)(x4−x2)(x3−x2)x1x3=x3(x3−x1)2+x4(x4−x2)2x3x4+(x3−x1)(x4−x2)(x3−x2)x1x4+(x3−x1)(x4−x2)(x3−x2)(x2−x1)(x4−x3)x1x2x3x4⩾(x3−x1)2+(x4−x2)2x4+(x3−x1)(x4−x2)(x3−x2)x1x4+(x3−x1)(x4−x2)(x3−x2)(x2−x1)(x4−x3)x1x2x3x4⩾∣(x3−x1)(x4−x2)∣x4. [2−∣x3−x2x1∣−∣(x3−x2)(x2−x1)(x4−x3)x1x3x2∣]⩾0.  \begin{array}{l} =\frac{\left(x_{1}-x_{3}\right)\left[\left(x_{3}-x_{2}\right)+\left(x_{1}-x_{3}\right)\right]}{x_{4}}- \\ \frac{\left(x_{2}-x_{3}\right)\left(x_{4}-x_{2}\right)}{x_{1}}+ \\ \frac{\left(x_{3}-x_{1}\right)\left[\left(x_{2}-x_{4}\right)+\left(x_{3}-x_{2}\right)\right]}{x_{2}}+ \\ \frac{\left(x_{4}-x_{2}\right)\left[\left(x_{4}-x_{2}\right)+\left(x_{2}-x_{3}\right)+\left(x_{3}-x_{1}\right)\right]}{x_{3}} \\ =\frac{\left(x_{3}-x_{1}\right)^{2}}{x_{4}}+\frac{\left(x_{4}-x_{2}\right)^{2}}{x_{3}}+ \\ \frac{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{3}-x_{2}\right)}{x_{2} x_{4}}- \\ \frac{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{3}-x_{2}\right)}{x_{2} x_{3}}+ \\ \frac{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{3}-x_{2}\right)}{x_{1} x_{3}} \\ =\frac{x_{3}\left(x_{3}-x_{1}\right)^{2}+x_{4}\left(x_{4}-x_{2}\right)^{2}}{x_{3} x_{4}}+ \\ \frac{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{3}-x_{2}\right)}{x_{1} x_{4}}+ \\ \frac{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{3}-x_{2}\right)\left(x_{2}-x_{1}\right)\left(x_{4}-x_{3}\right)}{x_{1} x_{2} x_{3} x_{4}} \\ \geqslant \frac{\left(x_{3}-x_{1}\right)^{2}+\left(x_{4}-x_{2}\right)^{2}}{x_{4}}+ \\ \frac{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{3}-x_{2}\right)}{x_{1} x_{4}}+ \\ \frac{\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\left(x_{3}-x_{2}\right)\left(x_{2}-x_{1}\right)\left(x_{4}-x_{3}\right)}{x_{1} x_{2} x_{3} x_{4}} \\ \geqslant \frac{\left|\left(x_{3}-x_{1}\right)\left(x_{4}-x_{2}\right)\right|}{x_{4}} \text {. } \\ {\left[2-\left|\frac{x_{3}-x_{2}}{x_{1}}\right|-\left|\frac{\left(x_{3}-x_{2}\right)\left(x_{2}-x_{1}\right)\left(x_{4}-x_{3}\right)}{x_{1} x_{3} x_{2}}\right|\right]} \\ \geqslant 0 \text {. } \\ \end{array}

When x1=x3,x2=x4x_{1}=x_{3}, x_{2}=x_{4}, i.e., a=c,b=da=c, b=d, the equality holds.

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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.