Question

Let x, yz be positive real numbers. Then prove that the following inequalities.
Exy S(x+y)(y+z)(2+2)
2. xy + y2 + 2xSx+y+z
x+y+z
nzs
x+y+z
x² + y² + 2
3
wyz(x+y+z) Sr'y? + y2 +22
​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.