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Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge \frac{5}{6}\cdot\sqrt{a^2+b^2+c^2+7}.$$ Here is what I've done so far.

By C-S $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{9}{2(a+b+c)}.$$ We need to prove $$\frac{9}{2(a+b+c)} \ge \frac{5}{6}\cdot\sqrt{a^2+b^2+c^2+7}=\frac{5}{6}\cdot\sqrt{(a+b+c)^2+5}.$$ Let $\sqrt{3}\le t,$ it becomes $$\frac{9}{2t}\ge \frac{5}{6}\sqrt{t^2+5} \iff \sqrt{3}\le t\le \sqrt{\frac{\sqrt{3541}-25}{10}}.$$ But I don't know how to prove it's true when $t> \sqrt{\dfrac{\sqrt{3541}-25}{10}}.$

I hope you help me prove inequality in this case. Also, if you have any ideas, you can share it.

Thank you.

Anonymous
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1 Answers1

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Proof.

The original is equivalent to $$\frac{a^2+b^2+c^2+3}{(a+b)(b+c)(c+a)}\ge \frac{5}{6}\cdot\sqrt{a^2+b^2+c^2+7}.$$ Notice that$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc\le a+b+c=\sqrt{a^2+b^2+c^2+2}.$

Let $1=ab+bc+ca\le x=a^2+b^2+c^2,$ it is $$\frac{x+3}{\sqrt{x+2}}\ge \frac{5}{6}\sqrt{x+7}\iff (x-2)\left(x+\frac{13}{11}\right)\ge 0 \iff x\ge 2.$$ If $1\le x\le 2$ we can use $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge \frac{5}{2}\ge \frac{5}{6}\cdot\sqrt{a^2+b^2+c^2+7}.$$ See also How can they know what type of inequality to use?

Hence, the proof is done.

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