Given three positive numbers a,b,c satisfying $$a^2+b^2+c^2=1$$ Prove that: $$\frac{bc}{a^2+1}+\frac{ac}{b^2+1}+\frac{ab}{c^2+1}\leq \frac{3}{4}$$ The things I have done so far: $$\sum \limits_{cyc}\frac{bc}{a^2+1}=\sum \limits_{cyc}\frac{bc}{2a^2+b^2+c^2}\leq \sum \limits_{cyc}\frac{bc}{2ab+2ac}$$ $$=\sum \limits_{cyc}\frac{bc}{2a(b+c)}\leq \frac{1}{4}\sum \limits_{cyc}\frac{(b+c)^2}{2a(b+c)}=\frac{1}{4}\sum \limits_{cyc}\frac{b+c}{2a}$$ $$=\frac{1}{8}.\frac{\sum \limits_{cyc}bc(b+c)}{abc}=\frac{1}{8}.\frac{\sum \limits_{cyc}bc(b+c)+3abc}{abc}-\frac{3}{8}$$ $$=\frac{1}{8}.\frac{(a+b+c)(ab+bc+ca)}{abc}-\frac{3}{8}$$ $$\leq \frac{1}{8}.\frac{\sqrt{3(a^2+b^2+c^2)}(a^2+b^2+c^2)}{abc}-\frac{3}{8}$$ $$=\frac{\sqrt{3}}{8abc}-\frac{3}{8}$$ I don't know what to do anymore.
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Michael Rozenberg
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Shizumi Aoki
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I would use that $\frac{bc}{a^2+1}\le \frac{2(b^2+c^2)}{1+a^2}=\frac{2(1-a^2)}{1+a^2}$ and so on. – Dr. Sonnhard Graubner Aug 07 '18 at 17:14
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Approach0 finds a dupe immediately. Downvote to any "trusted" user who didn't search. – Jyrki Lahtonen Aug 11 '18 at 17:48
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By AM-GM and C-S we obtain: $$\sum_{cyc}\frac{bc}{a^2+1}=\sum_{cyc}\frac{bc}{2a^2+b^2+c^2}\leq\frac{1}{4}\sum_{cyc}\frac{(b+c)^2}{a^2+b^2+a^2+c^2}\leq$$ $$\leq\frac{1}{4}\sum_{cyc}\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{b^2}{a^2+b^2}+\frac{a^2}{b^2+a^2}\right)=\frac{3}{4}.$$
Michael Rozenberg
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3How do you get $$ \frac{(b+c)^2}{a^2+b^2+a^2+c^2}\leq \left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right) , ? $$ – Diger Aug 07 '18 at 18:40
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Sorry. Can you elaborate a bit? The only thing I come up with is w.l.g. $c<b$ thus $$ \frac{(b+c)^2}{a^2+b^2+a^2+c^2}\leq \frac{2(b^2+c^2)}{2(a^2+c^2)} =\left(\frac{b^2}{a^2+c^2}+\frac{c^2}{a^2+c^2}\right) $$ – Diger Aug 07 '18 at 19:35
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@Diger Just by C-S $(a^2+b^2+a^2+c^2)\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)\geq(b+c)^2.$ – Michael Rozenberg Aug 07 '18 at 19:37
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