\frac{a+b+c}{2} = \sqrt a + \sqrt {b-1} + \sqrt {c-2}
Find (a+b)^c
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\frac{a+1}{2}\geq \sqrt{a}
\frac{(b-1)+1}{2}\geq \sqrt{b-1}
\frac{(c-2)+1}{2}\geq \sqrt{c-2}
adding these up we get:
\frac{a+b+c}{2} \geq \sqrt{a}+\sqrt{b-1}+\sqrt{c-2}
but we know we have the equality case, so a = 1, b-1 = 1, c-2 = 1 meaning that (a,b,c)=(1,2,3) so (a+b)^c=(1+2)^3= 27