Geometric Inequality: Prove √(2x)+√(2y)+√(2z)≤√(x+y)+√(y+z)+√(x+z)

[dengulakungen]

Homework Statement

Let a,b and c be lengths of sides in a triangle, show that
√(a+b-c)+√(a-b+c)+√(-a+b+c)≤√a+√b+√c

The Attempt at a Solution

With Ravi-transformation the expressions can be written as

√(2x)+√(2y)+√(2z)≤√(x+y)+√(y+z)+√(x+z).

Im stuck with this inequality. Can´t find a way to use any known inequalities such as AM-GM or the rearrangement inequality.

 

[QuantumQuest]

Have you thought about squaring?

 

[dengulakungen]

Yes I have tried to square the experisions, but without success. I will try it again.

 

[Ray Vickson]

Yes I have tried to square the experisions, but without success. I will try it again.

Your alternative expression of the inequality is the way to go. Even simpler: verify that for any two numbers $x,y>0$ we have $\sqrt{x+y} \geq \frac{1}{2} \sqrt{2x} + \frac{1}{2} \sqrt{2y}$. Again, have you tried squaring?