#### To determine

**To prove:** ex+yxy≥e2.

**Solution:** We proved the inequality as shown below.

#### Explanation

**Given:**

**Formulae used: **Calculate extreme values of function by equating partial derivative of the function to zero.

fx'(x,y)=0fy'(x,y)=0

To figure out whether the extreme points are minimum of maximum.

fxx(x,y)fyy(x,y)−f2xy(x,y)>0

or

fxx(x,y)fyy(x,y)−f2xy(x,y)<0fxx(x,y)fyy(x,y)−f2xy(x,y)=0

Consider f(x,y)=ex+yxy

Hence, use the formulae of extreme value of function of two variable

Now according to the given expression, to find the partial derivative of f(x,y)

fx(x,y)=ex+y(xy)−yex+y(xy)2fy(x,y)=ex+y(xy)−xex+y(xy)2fxx(x,y)=ex+y(xy)2−2y2(x−1)ex+y(xy)3fyy(x,y)=ex+y(xy)2−2x2(y−1)ex+y(xy)3

And

fx(x,y)=ex+y(xy)−yex+y(xy)2fxy(x,y)=ex+y(xy−x−y+1)(xy)2

The critical points are

fx(x,y)=0ex+y(xy)−yex+y(xy)2=0yex+y(x−1)=0

And

f y ( x,y )=0 e x+y ( xy )−x e x+y ( xy ) 2 =0 x e x+y ( y−1 )=0

By subtraction get:

yex+y(x−1)−xex+y(y−1)=0ex+y(xy−y−xy+x)=0ex+y(x−y)=0x=yxex+y(x−1)=0

Since ex+y≠0

Therefore, the critical points are (0,0) and (1,1)

Therefore, the condition of partial function f(x,y) to be minimum is

fxx(x,y)fyy(x,y)−f2xy(x,y)>0 and fxx(x,y)>0

The condition for minima at (1,1)

fxx(x,y)fyy(x,y)−f2xy(x,y)>0(e2)(e2)−0>0e4>0

And at (1,1)

fxx(x,y)>0e2>0

Since at (0,0) is neither maximum nor minimum.

Thus, the minimum value of the function f(x,y)=ex+yxy

f(x,y)=ex+yxyf(x,y)=e2≥0

Hence for every value when x,y∉(0,0)

ex+yxy≥e2