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Let $a,b,c$ are positive number such that $abc=1$. Prove that: $$\sqrt{a^2-a+1}+\sqrt{b^2-b+1}+\sqrt{c^2-c+1}\;\geq\; a+b+c$$

This problem froms my Math teacher. I have attempted to let $$(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$$. The inequality is equivalent to: $\frac{\sqrt{x^2-xy+y^2}}{y}+\frac{\sqrt{y^2-yz+z^2}}{z}+\frac{\sqrt{x^2-xz+z^2}}{x}\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}$

Then, I tried to use AM-GM but I stucked on it.

Michael Rozenberg
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Analyn_a
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  • I solved your problem. If you want to see my solution, show please your attempts. – Michael Rozenberg Aug 14 '19 at 11:55
  • I have attempted to let $(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$ or $(\frac{x^2}{yz}, \frac{y^2}{xz}, \frac{z^2}{xy})$ But I don't know what I will do next. – Analyn_a Aug 14 '19 at 11:57
  • Show it please. – Michael Rozenberg Aug 14 '19 at 11:59
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    Let $(a,b,c)=(\frac{x}{y}, \frac{y}{z}, \frac{z}{x})$. The inequality is equivalent to: $\sum \frac{\sqrt{x^2-xy+y^2}}{y}\geq \sum \frac{x}{y}$ $\sqrt{x^2-xy+y^2}=\sqrt{\frac{3}{4}(x+y)^2+\frac{1}{4}(x+y)^2}\geq \frac{1}{2}(x+y)$ $\Rightarrow \sum \frac{\sqrt{x^2-xy+y^2}}{y}\geq \sum \frac{x+y}{2y}$.

    But $\sum \frac{x+y}{2y}\leq \sum \frac{x}{y}$-> wrong.

    Then, I tried to use inequality $\sqrt{x^2-xy+y^2}\geq \frac{x^2+y^2}{x+y}$ but I don't know how to prove $\sum \frac{x^2+y^2}{x+y}\geq \sum \frac{x}{y}$ ...........

     – Analyn_a Aug 14 '19 at 12:09
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    You can keep squaring until there are integer index both sides and using the Müirhead’s Inequality – MafPrivate Aug 14 '19 at 12:22
  • @Analyn_a: Please [edit] the question and add all relevant information there: Where does the problem come from? What have to tried to solve the problem? Where are you stuck? That information should be in the question itself, not in the comments. – Martin R Aug 14 '19 at 12:33

1 Answers1

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The hint:

Use the Mixing Variables method.

Indeed, we can use the beautiful Can's idea.

Since $$\prod\limits_{cyc}(a-1)^2=\prod_{cyc}((a-1)(b-1))\geq0,$$ we can assume that $$(a-1)(b-1)\geq0$$ or $$a+b\leq1+ab=1+\frac{1}{c}.$$

Thus, by C-S:

\begin{align} \sqrt{a^2-a+1}&+\sqrt{b^2-b+1}\\ &=\sqrt{a^2+b^2-a-b+2+2\sqrt{(a^2-a+1)(b^2-b+1)}}\\ &\geq\sqrt{a^2+b^2-a-b+2+2\sqrt{\left(\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\right)\left(\left(b-\frac{1}{2}\right)^2+\frac{3}{4}\right)}}\\ &\geq\sqrt{a^2+b^2-a-b+2+2\left(\left(a-\frac{1}{2}\right)\left(b-\frac{1}{2}\right)+\frac{3}{4}\right)}\\ &=\sqrt{a^2+b^2-a-b+2+2ab-a-b+2}\\ &=\sqrt{(a+b)^2-2(a+b)+4}. \end{align}

But $f(x)=\sqrt{x^2-2x+4}-x$ decreases, which says $$\sum_{cyc}(\sqrt{a^2-a+1}-a)\geq f(a+b)+\sqrt{c^2-c+1}-c\geq$$ $$\geq f\left(1+\frac{1}{c}\right)+\sqrt{c^2-c+1}-c=\sqrt{3+\frac{1}{c^2}}-1-\frac{1}{c}+\sqrt{c^2-c+1}-c.$$ Id est, it's enough to prove that: $$\sqrt{3+\frac{1}{c^2}}-1-\frac{1}{c}+\sqrt{c^2-c+1}-c\geq0$$ and the rest is smooth.

Can you end it now?

C.F.G
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Michael Rozenberg
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