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?
Prove that $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0\Rightarrow a+b+c =2(\sqrt {ab}+\sqrt {bc}+\sqrt {ca})$
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algebra-precalculus
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0i think you forgot the Dollar sign – 2017-01-27
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0Yeah. Sorry @Dr.Sonnhard Graubner – 2017-01-27
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2and what about the variables $$a,b,c$$? – 2017-01-27
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3If $a^{\frac {1}{4}}+b^{\frac {1}{4}}+c^{\frac {1}{4}}=0$, necessarily $a=b=c=0$. – 2017-01-27
2 Answers
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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.
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0@ Rohan, I have a question to ask for you? Will you please help?please @Rohan – 2017-01-28
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0@NeWtoN What question? I will gladly help you if I know. Is it related to this question? – 2017-01-28
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0The question is not related but I have posted here, http://math.stackexchange.com/questions/2106883/graph-third-quartle/2106959#2106959. The answer I got here and the one given in my book doesn't match – 2017-01-28
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Maybe you mean that $$2(\sqrt{ab}+\sqrt{ac}+\sqrt{bc})-a-b-c=$$ $$=(\sqrt[4]a+\sqrt[4]b+\sqrt[4]c)(\sqrt[4]a+\sqrt[4]b-\sqrt[4]c)(\sqrt[4]a-\sqrt[4]b+\sqrt[4]c)(-\sqrt[4]a+\sqrt[4]b+\sqrt[4]c),$$ which is not necessary.