Question

plz do this sum......

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Answer 1

Step-by-step explanation:

Using Mathematica, we find

(x+y+z)3-x3-y3-z3=3(x+y)(y+z)(z+x)

(x+y+z)5-x5-y5-z5=5(x+y)(y+z)(z+x)(x2+y2+z2+xy+yz+zx)

Then we have to show that

30|(x+y)(y+z)(z+x)|<=45|(x+y)(y+z)(z+x)(x2+y2+z2+xy+yz+zx)|

If (x+y)(y+z)(z+x)=0 (that is, x+y=0,z=1 or y+z=0, x=1 or z+x=0, y=1), then we have the equality.

If (x+y)(y+z)(z+x)<>0, then the inequality becomes

3|x2+y2+z2+xy+yz+zx| >= 2 or, equivalently,

3((x+y)2+(y+z)2+(z+x)2)>= 4=((x+y)+(y+z)+(z+x))2,

which is true by the Cauchy-Schwarz inequality. The equality holds iff x+y=y+z=z+x, that is, iff x=y=z=1/3.

maybe this ans worg then report