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Ff $a,b,c$ are positive real numbers that $a^2+b^2+c^2=1$ ,Prove: $$\frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$$ Additional info: I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done: I tried to change LHS to something more easy to work but I was not successful. For example $$\frac{ab}{1+c^2}=\frac{1}{2}\left(\frac{a^2+b^2+2ab}{1+c^2}-\frac{a^2+b^2}{1+c^2}\right)$$

that was not useful. Any hint for starting step is appreciated.

user2838619
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2 Answers2

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Write $x=a^2$, $y=b^2$, and $z=c^2$. Then, $x+y+z=1$. A first consequence of this is $$ (1+z)\geq 2\sqrt{xy+z}.\tag{*} $$ This is true because $$ (1+z)^2-(2\sqrt{xy+z})^2=(1+z)^2-4z-4xy\\ =(1-z)^2-4xy=(x+y)^2-4xy=(x-y)^2\geq 0. $$ Analogous to (*), we also have $$ (1+y)\geq2\sqrt{xz+y},\quad (1+x)\geq 2\sqrt{yz+x}. $$ This implies $$ \sum_{\text{cyc}}\frac{ab}{1+c^2}=\sum_{\text{cyc}}\frac{\sqrt{xy}}{1+z}\leq\frac{1}{2}\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}. $$ So the claim follows if we can show $\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}\leq\frac{3}{2}$. But that has already been done here, so we are done!

p.s. For completeness, I'll produce the argument from the link here: first, observe $$ xy+z=xy+(1-x-y)=(1-x)(1-y)=(y+z)(x+z). $$ Then, it follows from the AM-GM inequality that $$ \sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{xy+z}}=\sum_{\text{cyc}}\frac{\sqrt{xy}}{\sqrt{(x+z)(y+z)}}\leq\sum_{\text{cyc}}\frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{3}{2}. $$

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Kim Jong Un
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$\displaystyle \frac{ab}{1+c^2}+\frac{bc}{1+a^2}+\frac{ca}{1+b^2}\le\frac{3}{4}$ is equivalent to

EDIT: Do not pursue this method. It is wrong! Left as warning.

$\displaystyle \frac{a^2+b^2}{1+c^2}+\frac{b^2+c^2}{1+a^2}+\frac{c^2+a^2}{1+b^2}\le\frac{3}{2}$, as $ab\leq\frac{1}{2}(a^2+b^2)$.

This is equivalent to:

$\displaystyle \frac{1-c^2}{1+c^2}+\frac{1-b^2}{1+b^2}+\frac{1-a^2}{1+a^2} \leq \frac{3}{2}$, where $a^2+b^2+c^2=1$.

Let $x=a^2,y=b^2,z=c^2$, for simplicity.

Then the above is equivalent to

$\displaystyle \frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z} \leq \frac{9}{4}$, whre $x+y+z=1$.

ShakesBeer
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  • Similarly, writing $1+c^2=a^2+b^2+2c^2\geq 2(ab+c^2)$ (and similarly for the others) led to a dead end for me. $\sum_\text{cyc}\frac{ab}{ab+c^2}$ can actually be greater than $\frac{3}{2}$. – Kim Jong Un Aug 29 '14 at 10:40
  • In fact, it is easy to show that the reverse is always true, i.e. $$\sum_{cyc} \frac{1-a^2}{1+a^2} \ge \frac32$$ by cyclically summing $$f(t) = \frac{1-t^2}{1+t^2}-\frac12-\frac98\left(\frac{1}{3}-t^2\right) = \frac{\left(-1+3 t^2\right)^2}{8 \left(1+t^2\right)} \ge 0$$ – Macavity Aug 29 '14 at 13:05
  • Yup, I noticed that :D – ShakesBeer Aug 29 '14 at 13:26