9take
a+b-c = x
and so on
x + y + z =a + b + c
√x + ...... ≤ √x+y /2 .....
proceed now
If a,b,c are sides of a triangle, prove that
\sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} \le \sqrt{a} + \sqrt{b} + \sqrt{c}
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3 Answers
91take
a+b-c=x
b+c-a=y
c+a-b=z
so, we have to prove that :
\sqrt{x}+\sqrt{y}+\sqrt{z} \leq \sqrt{\frac{x+y}{2}}+\sqrt{\frac{y+z}{2}}+\sqrt{\frac{z+x}{2}}
now for x>0, the function \sqrt{x} is concave.
so, by jensen's we have :
\frac{\sqrt{x}+\sqrt{y}}{2} \leq \sqrt{\frac{x+y}{2}}
\frac{\sqrt{y}+\sqrt{z}}{2} \leq \sqrt{\frac{y+z}{2}}
\frac{\sqrt{z}+\sqrt{x}}{2} \leq \sqrt{\frac{z+x}{2}}
adding the three equations we get :
\sqrt{x}+\sqrt{y}+\sqrt{z} \leq \sqrt{\frac{x+y}{2}}+\sqrt{\frac{y+z}{2}}+\sqrt{\frac{z+x}{2}}
