Maximum value

Algebra Level 5

Let x , y , z x,y,z are non-zero finite real numbers. Now, consider the following expression S = ( x + y − z ) 2 ( x + y ) 2 + z 2 + ( y + z − x ) 2 ( y + z ) 2 + x 2 + ( z + x − y ) 2 ( z + x ) 2 + y 2 S=\frac{(x+y-z)^2}{(x+y)^2+z^2}+\frac{(y+z-x)^2}{(y+z)^2+x^2}+\frac{(z+x-y)^2}{(z+x)^2+y^2}

Let, the maximum value of S S is S 0 S_0 and it occurs at x = x 0 , y = y 0 , z = z 0 x=x_0,\, y=y_0,\, z=z_0 .

Find the value of ( S 0 + x 0 + y 0 + z 0 ) \big(S_0+x_0+y_0+z_0\big) .


The answer is 6.

Anirban Karan by Anirban Karan , 30, India

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2 solutions

Anirban Karan
Apr 22, 2017

( x + y − z ) 2 ( x + y ) 2 + z 2 = ( x + y − z ) 2 + ( x + y + z ) 2 ( x + y ) 2 + z 2 − ( x + y + z ) 2 ( x + y ) 2 + z 2 = 2 [ ( x + y ) 2 + z 2 ] ( x + y ) 2 + z 2 − ( x + y + z ) 2 ( x + y ) 2 + z 2 = 2 − ( x + y + z ) 2 ( x + y ) 2 + z 2 \begin{aligned}\frac{(x+y-z)^2}{(x+y)^2+z^2}&=\frac{(x+y-z)^2+(x+y+z)^2}{(x+y)^2+z^2}-\frac{(x+y+z)^2}{(x+y)^2+z^2}&\\ &=\frac{2[(x+y)^2+z^2]}{(x+y)^2+z^2}-\frac{(x+y+z)^2}{(x+y)^2+z^2}&\\ &=2-\frac{(x+y+z)^2}{(x+y)^2+z^2}\end{aligned} ⟹ S = 6 − ( x + y + z ) 2 [ 1 ( x + y ) 2 + z 2 + 1 ( y + z ) 2 + x 2 + 1 ( z + x ) 2 + y 2 ] \implies S=6-(x+y+z)^2\Bigg[\frac{1}{(x+y)^2+z^2}+\frac{1}{(y+z)^2+x^2}+\frac{1}{(z+x)^2+y^2}\Bigg]

Now, let T = ( x + y + z ) 2 [ 1 ( x + y ) 2 + z 2 + 1 ( y + z ) 2 + x 2 + 1 ( z + x ) 2 + y 2 ] \displaystyle T=(x+y+z)^2\Bigg[\frac{1}{(x+y)^2+z^2}+\frac{1}{(y+z)^2+x^2}+\frac{1}{(z+x)^2+y^2}\Bigg] .

Then S S is maximum if T T is minimum.

Again, as x , y , z x,y,z are non-zero finite real numbers and T T contains sum of squares, so T ≥ 0 T\geq 0 .

Hence, the minimum value of T T is 0 and it occurs for x 0 + y 0 + z 0 = 0 x_0+y_0+z_0=0 .

Thus, S 0 = 6 S_0=6 and S 0 + x 0 + y 0 + z 0 = 6 \boxed{S_0+x_0+y_0+z_0=6}

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Frank Petiprin
Oct 10, 2017 
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import numpy as np
#Pythonista Program ran on Ipad 2

def findval(a,b,c):
    val1=(a+b-c)**2/((a+b)**2+c**2)
    val2=(b+c-a)**2/((b+c)**2+a**2)
    val3=(c+a-b)**2/((c+a)**2+b**2)
    val=val1+val2+val3
    return val
#: starter for first max found
max=-10000
#: Increment within intervals can be varied. 1.00,  0.5, 0.125, 0.0625 
incr=.25
#: Interval-1 [-10,10] can vary 
slef1,srig1=-10,10
#: Interval-2 [-10,10]
slef2,srig2=-10,10
#:Interval-3 [-10,10]
slef3,srig3=-10,10
lef3=slef3
while(lef3<=srig3):
    lef2=slef2
    while (lef2<=srig2):
      lef1=slef1
      while (lef1<=srig1):
        x,y,z=lef3,lef2,lef1  
        if(((x!=0)and(y!=0)and(z!=0))):
          Sval=findval(x,y,z)
          if(Sval>max):
            max=Sval
            sax,say,saz=x,y,z
        lef1=lef1+incr
      lef2=lef2+incr
    lef3=lef3+incr
print('three x,y,z intervals are the same,','[',slef1,',',srig1,']')
print('increment on intervals = ',incr)
print('maximum =', max,'x y z', sax, say, saz)
print('halt++++++++++++++++++++++++++++++++++++++')
#THREE PROGRAM RUNS

the three x,y,z intervals are the same, [ -1 , 1 ]
increment on the intervals =  0.25
maximum = 6.0. x  y  z  -1  0.25.  0.75

halt++++++++++++++++++++++++++++++++++++++

the three x,y,z intervals are the same, [ -5 , 5 ]
increment on the intervals =  0.25
maximum = 6.0  x y z  -5  0.25  4.75

halt++++++++++++++++++++++++++++++++++++++

the three x,y,z intervals are the same, [ -10 , 10 ]
increment on the intervals =  0.25
maximum = 6.0  x  y  z   -10   0.25.  9.75

halt++++++++++++++++++++++++++++++++++++++

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