prove exey2 ex2y2Solutionexey2 ex2y2 Divide both sides by

prove (e^x+e^y)/2 >= e^(x/2+y/2)

(e^x+e^y)/2 >= e^(x/2+y/2)

Divide both sides by e^(x/2+y/2)

e^(x/2 -y/2) + e^(y/2 -x/2) >=2

Let x/2 -y/2 = z

e^z +e^-z >=2

If we prove the above inequality then our proof is done

Now for e^z + e^-z minimum occurs at z =0 i.e. e^0 +e^0 = 2

f(z) = e^z +e^-z : f\'(z) = e^z -e^-z

f\'(z) = 0 ; e^z = e^-z ----> z =0

f\"(z) = e^z +e^-z at z = 0 ; f\"(z) =2

Hence f\"(z) >0 so at z =0 f(z) has minima value which is 2

Hence f(z) is always greater than 2

So, (e^x+e^y)/2 >= e^(x/2+y/2)