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let $a,b,c>0$, such $ab+bc+ac=1$,show that $$\dfrac{1}{3a+5b+7c}+\dfrac{1}{3b+5c+7a}+\dfrac{1}{3c+5a+7b}\le\dfrac{\sqrt{3}}{4}$$

by Macavity C-S:with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$ $$\dfrac{1}{3a+5b+7a}\le\dfrac{1}{2}\left(\dfrac{1}{6a+9b}+\dfrac{1}{b+7c}\right)$$ It suffices to show $$\sum_{cyc}\left(\dfrac{1}{6a+9b}+\dfrac{1}{b+7c}\right)\le\dfrac{\sqrt{3}}{2}$$

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Using Cauchy-Schwarz in a different way: $$\begin{align} 2(3a+5b+7c) &= 15(a+b+c)-(9a+5b+c) \\ &= \sqrt{(118+107)(2+a^2+b^2+c^2)} - (9a+5b+c)\\ &\ge \sqrt{118\cdot2}+\sqrt{(9^2+5^2+1^2)(a^2+b^2+c^2)} -(9a+5b+c)\\ &\ge 2\sqrt{59} \end{align} $$

$$\implies \sum_{cyc} \frac1{3a+5b+7c} \le \frac3{\sqrt{59}} < \frac{\sqrt3}4$$

P.S. The maximum is in fact $\dfrac{\sqrt3}5$, though a simple way to show that eludes me for now.

Macavity
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  • can you post the maximum is $\dfrac{\sqrt{3}}{5}$? I can't prove it –  Jun 25 '15 at 16:34
  • The question about the maximum is now asked in http://math.stackexchange.com/questions/1339079/how-to-prove-this-maximum-is-frac-sqrt35/ – Jose Brox Jun 26 '15 at 00:19