$x$, $y$ and $z$ are lengths of the sides of a triangle. Prove the following and determine when both LHS and RHS are not an inequality but EQUAL.
$$\sqrt{x+y-z} + \sqrt{y+z-x} + \sqrt{z+x-y} \leqslant \sqrt{x}+\sqrt{y}+\sqrt{z}$$
Asked
Active
Viewed 91 times
1
-
At least one case of equality - probably the only one - should be obvious. – Joffan Jan 25 '17 at 15:53
-
use the Ravi-substitution – Dr. Sonnhard Graubner Jan 25 '17 at 15:53
1 Answers
3
using the Ravi-substitution $$x=b+c,y=a+c,z=a+b$$ we get $$\sqrt{2a}+\sqrt{2b}+\sqrt{2c}\le \sqrt {a+b}+\sqrt{a+c}+\sqrt{b+c}$$ but we have $$\sqrt{2a}+\sqrt{2b}+\sqrt{2c}=\frac{\sqrt{2a}+\sqrt{2b}}{2}+\frac{\sqrt{2a}+\sqrt{2c}}{2}+\frac{\sqrt{2b}+\sqrt{2c}}{2}\le\sqrt{\frac{2a+2b}{2}}+\sqrt{\frac{2a+2c}{2}}+\sqrt{\frac{2c+2b}{2}}=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}$$
Dr. Sonnhard Graubner
- 97,058