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Let $a,b,c>0: a+b+c=\dfrac{1}{a}+ \dfrac{1}{b}+ \dfrac{1}{c}.$ Prove that \begin{align*} \frac{a}{b} +\frac{b}{c}+ \frac{c}{a} \ge \sqrt{4\left(a^2+b^2+c^2 \right) -ab-bc-ca} \end{align*}

I tried to square but I did not find any interesting point.

Hope to see some ideas. Thank you.

Updated: We will prove \begin{align*} \frac{a}{b} +\frac{b}{c}+ \frac{c}{a} \ge \sqrt{\left(\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right)\frac{4\left(a^2+b^2+c^2 \right) -ab-bc-ca}{a+b+c} } \end{align*} By computer, I got the equivalent form $$\sum_{\text{cyc}} a \left(\frac{a}{b} -\frac{b}{c} \right) \left(\frac{a}{b}- \frac{b}{c} -3 \right) \ge 0.$$ From here, I don't know how to continue Just notice that, equality occurs at $a=b=c>0,$ also for $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=6,$$ and $$\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=9.$$

Dragon boy
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1 Answers1

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Remark: Here is a proof which is motivated by computer.

We need to prove that $$\left(\frac{a}{b} +\frac{b}{c}+ \frac{c}{a}\right)^2 \ge \left(\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c} \right)\frac{4\left(a^2+b^2+c^2 \right) -ab-bc-ca}{a+b+c}$$ or \begin{align*} &{a}^{5}{c}^{2}-2\,{a}^{4}{b}^{2}c-3\,{a}^{4}b{c}^{2}+{a}^{4}{c}^{3}+{a }^{3}{b}^{4}+3\,{a}^{3}{b}^{3}c+3\,{a}^{3}b{c}^{3}+{a}^{2}{b}^{5}\\[6pt] &\qquad -3\,{ a}^{2}{b}^{4}c-2\,{a}^{2}b{c}^{4}-2\,a{b}^{4}{c}^{2}+3\,a{b}^{3}{c}^{3 }-3\,a{b}^{2}{c}^{4}+{b}^{3}{c}^{4}+{b}^{2}{c}^{5}\\[6pt] &\ge 0. \tag{1} \end{align*}

We have the following SOS (Sum of Squares): \begin{align*} &2(a + b + c)\cdot \mathrm{LHS}_{(1)}\\[6pt] ={}& \frac32\, \left( b{a}^{2}c-{a}^{2}{c}^{2}-a{b}^{3}+a{b}^{2}c-2\,ab{c}^{2}+ {b}^{2}{c}^{2}+b{c}^{3} \right) ^{2}\\ &\qquad + \frac12\, \left( 2\,{a}^{3}c-2\,{a}^{2}{b}^{2}-3\,b{a}^{2}c+{a}^{2}{c}^{2} -a{b}^{3}+3\,a{b}^{2}c+{b}^{2}{c}^{2}-b{c}^{3} \right) ^{2}\\[6pt] &\qquad + \sum_{\mathrm{cyc}} 2\,ab \left( {a}^{2}c-a{b}^{2}-abc-a{c}^{2}+2\,b{c}^{2} \right) ^{2}. \tag{2} \end{align*} Thus, (1) is true. We are done.

River Li
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