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$x$, $y$ 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.

$$\sqrt{x+y-z} + \sqrt{y+z-x} + \sqrt{z+x-y} \leqslant \sqrt{x}+\sqrt{y}+\sqrt{z}$$

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    At least one case of equality - probably the only one - should be obvious. – 2017-01-25
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    use the Ravi-substitution – 2017-01-25

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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 $$\sqrt{2a}+\sqrt{2b}+\sqrt{2c}=\frac{\sqrt{2a}+\sqrt{2b}}{2}+\frac{\sqrt{2a}+\sqrt{2c}}{2}+\frac{\sqrt{2b}+\sqrt{2c}}{2}\le\sqrt{\frac{2a+2b}{2}}+\sqrt{\frac{2a+2c}{2}}+\sqrt{\frac{2c+2b}{2}}=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}$$