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Let $a,b$ and $c$ be three real positive numbers. Prove that

$$\sum_{sym}\frac{1}{a^2+4b^2+9c^2}\le\frac{9}{14}(a^2+b^2+c^2)$$

I tried to use Cauchy-Schwarz :

$$\sum_{sym}\frac{1}{a^2+4b^2+9c^2}\le\frac{1}{9}\sum_{sym}\left(\frac{1}{a^2}+\frac{1}{4b^2}+\frac{1}{9c^2}\right)=\frac{49}{324}\sum_{sym}\frac{1}{a^2}$$ and therefore I need to prove that

$$\sum_{sym}\frac{1}{a^2}\le \frac{1458}{343}\sum_{sym}a^2$$ I'm not sure if the last inequality is true.

Thanks for any help.

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    $\begingroup$ this is not true. $a=b=c \implies a^4 \ge \dfrac{1}{9}$. if there is no other condition, you can't prove it. $\endgroup$ – chenbai Jul 17 '14 at 1:52
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    $\begingroup$ what does $\sum_{sym}$ mean? $\endgroup$ – Will Jagy Jul 17 '14 at 1:58
  • $\begingroup$ The inequality doesn't even look plausible. Letting $a,b,c\to0$ the left hand side blows up and the right hand side goes to zero. $\endgroup$ – user940 Jul 30 '15 at 16:08

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