# Prove $\frac {x+y}{x^2+y^2} + \frac {y+z}{y^2+z^2} + \frac {z+x}{z^2+x^2} \leq \frac 1x + \frac 1y + \frac 1z .$ [duplicate]

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, gebruikerMar 15 '16 at 10:04

• – colormegone Mar 15 '16 at 4:41

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 $\frac{y+z}{y^2+z^2} \le \frac{1}{y}$

And

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.

• How did you get the 2 on the LHS? $2(x+y)$ – CAGT Mar 15 '16 at 4:35
• 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. – thkim1011 Mar 15 '16 at 4:39