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Prove that if $x,y,z$ are positive real numbers then the following inequality holds $$ \frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2} \leq \frac 1x + \frac 1y + \frac 1z . $$

I tried to put all of them under a common denominator, but that didn't work. I don't know how to find something greater than the LHS and less than the RHS to prove the inequality.


marked as duplicate by Macavity, Martin R, Joel Reyes Noche, Harish Chandra Rajpoot, gebruiker Mar 15 '16 at 10:04

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This is an inequality with symmetry i.e. you can change $x$ by $y$ , $y$ by $z$ and $z$ by $x$ or in any other way replacing one variable by the other but the inequality always remains the same.

And this property of inequalities has a special advantage, i.e. we can assume an arbitrary order of magnitude for the variables (in case I didn't get the terminology right, I wanted to say that we can arbitrarily choose which variable is greater than which and so on.)

Hence, without any loss of generality, we can assume that $x \le y \le z$ where $x,y,z$ are all positive reals.

So we have $$x \le y$$ $$xy \le y^2$$ $$x^2+xy \le x^2+y^2$$ $$\frac{x+y}{x^2+y^2} \le \frac{1}{x}$$

Similarly, from $y \le z$, we can deduce that $\frac{y+z}{y^2+z^2} \le \frac{1}{y}$


From $x \le z$, we can deduce that $\frac{x+z}{x^2+z^2} \le \frac{1}{z}$

Hope this helps you.


Since $x^2+y^2\ge 2xy$, $$\frac{x+y}{x^2+y^2}\le \frac{x+y}{2xy}=\frac{1}{2x}+\frac{1}{2y}.$$ Similarly to the other factors and add them up.


I'm not too experienced with inequalities, but I managed to solve the problem, so I'll attempt to explain how I got to the solution.

In general, symmetric inequalities such as these seem to be breakable in my past experience (of course I haven't had much experience, so it would be nice if someone could verify that this is true). By that I mean that you are likely to be able to prove different inequalities that sum to what you want.

In this problem, I figured it would be nice to prove that $$2\left(\frac{x + y}{x^2 + y^2}\right) \le \frac1x + \frac 1y \iff 2(x + y) \le (x^2 + y^2)\left(\frac{1}{x} + \frac{1}{y}\right).$$ If this were true, then we can recreate the same inequalities in $y,z$ and $x,z$, and sum those to get what we want.

So now let's put our focus to proving this new simpler inequality. I expanded the right side to get $$ 2x + 2y \le x + y + \frac{x^2}{y} + \frac{y^2}{x} \implies 0 \le -x - y + \frac{x^2}{y} + \frac{y^2}{x}.$$ Now we have struck gold. The right side factor very nicely. We now have to show that $$0\le (y^2 - x^2)\left(\frac 1x - \frac 1y\right) = (y - x)(y + x)\left(\frac{y - x}{xy}\right) = \frac{(y-x)^2(y + x)}{xy}.$$ Every term in the fraction in the right side is positive, so this inequality is definitely true. Therefore, we have proven the inequality.

In order to write up a solution, just reverse the steps to start from scratch and culminate in the given inequality.

  • $\begingroup$ How did you get the 2 on the LHS? $2(x+y)$ $\endgroup$ – CAGT Mar 15 '16 at 4:35
  • 1
    $\begingroup$ I clarified the solution a little bit. Basically if you sum up the three inequality in $x,y$, $y,z$, and $x,z$, there will be $$\frac 2x + \frac 2y + \frac2z$$ on the right side. The $2$ is there to eliminate it. $\endgroup$ – thkim1011 Mar 15 '16 at 4:39

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