Prove for all a,b,c>0: a/(b+c) + b/(a+c) + c/(a+b) >= 3/2 ?

[barbiemathgurl]

can somebody prove that for all a,b,c>0:

a/(b+c) + b/(a+c) + c/(a+b) >= 3/2

 

[morphism]

There are probably a few ways you can do this.

Here's a hint: Set x=b+c, y=a+c, and z=a+b. Then re-write the left side as: 1/2 [(x/y + y/x) + (y/z + z/y) + (z/x + x/z) - 3].

 

[tacman]

To prove this inequality, we will use the Cauchy-Schwarz inequality, which states that for any real numbers x1, x2, ..., xn and y1, y2, ..., yn, we have:

(x1^2 + x2^2 + ... + xn^2)(y1^2 + y2^2 + ... + yn^2) >= (x1y1 + x2y2 + ... + xnyn)^2

First, we will rewrite the left side of the given inequality as:

a/(b+c) + b/(a+c) + c/(a+b) = (a^2 + b^2 + c^2)/(ab + ac + bc)

Now, let's consider the following set of numbers:

x1 = a
x2 = b
x3 = c
y1 = 1/(b+c)
y2 = 1/(a+c)
y3 = 1/(a+b)

Applying the Cauchy-Schwarz inequality, we get:

(a^2 + b^2 + c^2)(1/(b+c)^2 + 1/(a+c)^2 + 1/(a+b)^2) >= (a/(b+c) + b/(a+c) + c/(a+b))^2

Expanding the left side, we have:

(a^2 + b^2 + c^2)((1/(b+c))^2 + (1/(a+c))^2 + (1/(a+b))^2) >= (a/(b+c) + b/(a+c) + c/(a+b))^2

Simplifying, we get:

(a^2 + b^2 + c^2)(1/(b^2 + 2bc + c^2) + 1/(a^2 + 2ac + c^2) + 1/(a^2 + 2ab + b^2)) >= (a/(b+c) + b/(a+c) + c/(a+b))^2

Next, we can use the AM-GM inequality, which states that for any positive numbers x and y, we have:

(x + y)/2 >= sqrt(xy)

Applying this inequality to each of the denominators on the left side, we get:

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