Prove that $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0\Rightarrow a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$ If $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0$, prove that $a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$.
I guess that the given expression is true if and only if $a=b=c=0$. Is it true? Or,is there any other alternatives? 

We have $$a^{1/4} + b^{1/4} + c^{1/4} =0$$ $$\Rightarrow  (a^{1/4} + b^{1/4} + c^{1/4})^2 = a^{1/2}+b^{1/2} + c^{1/2} + 2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}]=0$$ $$\Rightarrow a^{1/2}+b^{1/2}+c^{1/2}=-2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}] $$ $$\Rightarrow (a^{1/2}+b^{1/2}+c^{1/2})^2 =(-2 [(ab)^{1/4}+(bc)^{1/4}+(ac)^{1/4}])^2$$ $$\Rightarrow a+b+c+2 [\sqrt{ab}+\sqrt {bc}+\sqrt {ac}] = 4 [\sqrt {ab} + \sqrt {bc} + \sqrt {ac} +2 [(a^2bc)^{1/4}+(ab^2c)^{1/4}+(abc^2)^{1/4}]] $$ $$\Rightarrow a+b+c-2 [\sqrt {ab}+\sqrt {bc}+\sqrt {ac}] =8 (abc)^{1/4}[a^{1/4}+b^{1/4}+c^{1/4}] =0$$ $$\boxed {a+b+c =2[\sqrt {ab}+\sqrt {bc}+\sqrt {ac}]}$$ Hope it helps.
