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If $a$, $b$, $c$ are nonzero natural numbers and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$, show that $a\sqrt{bc} + b\sqrt{ac} + c\sqrt{ab} \leq abc$. I basically have no idea how I can start this problem. I thought of demonstrating that $a\sqrt{bc}\leq abc$ but I just got lost in all the equations. Also, I thought to give a common factor but I don't really find a good point of starting the problem with. If you have any other idea, please write in the comments down below. Any idea is welcome! Hope one of you can help me! Thank you so much!

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    $\begingroup$ The claim is equivalent to $$\frac1{\sqrt{bc}}+\frac1{\sqrt{ac}}+\frac1{\sqrt{ab}}\le1.$$ AM-GM says $$\frac1{\sqrt{xy}}\le\frac12(\frac1x+\frac1y).$$ $\endgroup$ Commented Jan 5, 2023 at 18:21

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Multiply $abc$ in $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$ to get:$$bc+ac+ab=abc$$Now, by AM-GM: $$a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\le\frac12(a(b+c)+b(a+c)+c(a+b))=\frac12(ab+ac+ab+bc+ac+bc)=ab+bc+ac=abc$$

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