Tutor Hunt Questions

>>Right Hand Side:
The three term
F(a) = a/sqrt(a^2+b^2)
F(b) = b/sqrt(b^2+c^2)
F(c) = c/sqrt(c^2+a^2)
No matter how we order a, b and c, we can always have two of the terms F(a), F(b) and F(c) above or below 1/sqrt(2) at the same time.

Let`s assume the two terms are F(a) and F(b):
[F(a)-1/sqrt(2)]*[F(b)-1/sqrt(2)]>=0.

Therefore:
F(a)/sqrt(2)+F(b)/sqrt(2)
<=F(a)F(b)/2+1/2
<=ab/sqrt[(a^2+b^2)*(b^2+c^2)]+1/2
<=ab/(ab+bc)+1/2=a/(a+c)+1/2
=3/2-c/(a+c)<=3/2-c/sqrt(a^2+c^2)sqrt(2)

Therefore:
[F(a)+F(b)+F(c)]/sqrt(2)<=3/2

>>Left Hand Side:
F(a)+F(b)+F(c)
=1/sqrt(1+b^2/a^2)+1/sqrt(1+c^2/b^2)+1/sqrt(1+a^2/c^2).
Let x=b^2/a^2, y=c^2/b^2 and z=a^2/c^2

Using xyz=1, we have:
F(a)+F(b)+F(c)
=1/sqrt(1+x)+1/sqrt(1+y)+1/sqrt(1+z)
>=1/(1+x)+1/(1+y)+1/(1+z)
=(3+2x+2y+2z+xy+xz+yz)/[(1+x)(1+y)(1+z)]
>=(1+xyz+x+y+z+xy+xz+yz)/[(1+x)(1+y)(1+z)]
=1

As Ojuade wrote its straightforward, choose a=b=c=1/2
you will get the right hand side equal to 1.5, and 1<1.5.
3/sqrt(2)=2.1232 >1.5.

hope this help.

This is pretty straightforward.
Positive real numbers are 1,2,3,4,5....
Now, let a=1, b=2, c=3 and substituting these into the equation

a/Sqrt(a^2+b^2)+b/Sqrt(b^2+c^2)+c/Sqrt(c^2+a^2)

= 1/√(5) + 2/√(13) + 3/√(10) = 1.95

This clearly shows that:

1<1.95
and
3/√(2) = 2.12 which is > 1.95

Suppose that a>0, b> and c>0.

Now for each term:

a/Sqrt(a^2+b^2) has a solution between (0,1),
b/Sqrt(b^2+c^2) has a solution between (0,1),
c/Sqrt(c^2+a^2) has a solution between (0,1).

Taking the limit as a, b and c to reach infinity, we obtain:

a/Sqrt(a^2+b^2) approaches cos(pi/4)=1/sqrt(2),
b/Sqrt(b^2+c^2) approaches cos(pi/4)=1/sqrt(2),
c/Sqrt(c^2+a^2) approaches cos(pi/4)=1/sqrt(2).

Summing up all three terms above, we get an equation which satisfies:

a/Sqrt(a^2+b^2)+b/Sqrt(b^2+c^2)+c/Sqrt(c^2+a^2)<3/sqrt(2), where:

1/sqrt(2)+1/sqrt(2)+1/sqrt(2)=3/sqrt(2).

Hope this helps :)