Careful with AM-GM!

Algebra Level 5

Let x 1 , x 2 , x 3 , x 4 x_1, x_2, x_3, x_4 be distinct real numbers such that x 1 x 3 = x 2 x 4 x_1 x_3 = x_2 x_4 and x 1 x 2 + x 2 x 3 + x 3 x 4 + x 4 x 1 = 4 \frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_4}+\frac{x_4}{x_1}=4

Find the maximum value of x 1 x 3 + x 2 x 4 + x 3 x 1 + x 4 x 2 \frac{x_1}{x_3}+\frac{x_2}{x_4}+\frac{x_3}{x_1}+\frac{x_4}{x_2} .


The answer is -12.

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1 solution

Kushal Dey
Dec 30, 2020

Since we are given that x1x3=x2x4 we can take x1/x2=x4/x3=a and x2/x3=x1/x4=b. Then it is not difficult to see that our given equation is a+1/a+b+1/b=4, and the expression we want to evaluate is (a+1/a)(b+1/b). Let's take b+1/b=-c. The reason taking a variable like that is if a is positive then a+1/a>=2 via AM-GM inequality, however if a+1/a<4 then that would make 0<b+1/b<=2. LHS of this inequality is okay, but the RHS is not completely okay(due to AM-GM inequality). Thus if a,b are both positive then a=b=1. But that case would make x1=x2=x3=x4 which is a contradiction to the condition given in the question. Thus one of them must be negative and suppose we take b as our negative entry. Thus b+1/b<=-2 and that makes c>=2 and a+1/a=4+c. Thus our required expression now becomes-c(4+c)=4-(c+2)² Max value is clearly attained at c=2(c cannot be -2 since we argued thatc>=2) which is -12. And it is attained at (t,ut,-ut,-t) for all real t and u is a root of the quadratic u²-6u+1=0.

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