BdMO Online Forum

Let $x,y,z$ be positive real numbers such that $xy+yz+zx=3xyz$ . Prove that :

$$x^2y+y^2z+z^2x \ge 2(x+y+z)-3$$

The thing that always bums me out in ineq probs is the weird constant terms (often +1,-1,+3,-3) . in this prob, its easy to get rid of that; $3 = \frac1x + \frac1y + \frac1z $ . Of course, minus signs never help, so we move these to the left.

At this point, you will probably be bothered by the annoying coefficient of $(x+y+z)$. I mean, $2$ is such an annoying thing to have in a three variable inequality. So we add $x+y+z$ to both sides . Now our inequality looks like this.

$$x^2y+y^2z+z^2x+\frac1x + \frac1y + \frac1z +x+y+z \ge 3(x+y+z)$$ .

There are $9$ terms on the left, and a coefficient of $3$ on the right. This indicates that we should apply AM-GM in groups of three. By picking $x^2y+x+1/y\ge 3x$ cyclically and summing them up, we end up with the desired inequality.

I think 2 is not annoying here as you can simply have by AM-GM,
\[ \sum_{cyc} (x^2y+\dfrac{1}{y}) \geq 2\sum_{cyc} x \]
the desired inequality!!!

Hi, I’m new here but im trying this exact question from 2017 PAMO!

I know the thread is years old, but can anyone help me understand this am—gm and how these statements are true? I get the typical am-gm like( x1+x2+x3+…+xn/n)>=nth root(x1x2x3…xn), but x²y+1/y? Looks completely different. I’m very new to this stuff so if you say “read this book” thats fine

Levid0 wrote: ↑

Fri May 28, 2021 3:41 am

Hi, I’m new here but im trying this exact question from 2017 PAMO!

I know the thread is years old, but can anyone help me understand this am—gm and how these statements are true? I get the typical am-gm like( x1+x2+x3+…+xn/n)>=nth root(x1x2x3…xn), but x²y+1/y? Looks completely different. I’m very new to this stuff so if you say “read this book” thats fine

\[\dfrac{x^2y+\frac{1}{y}}{2}\geq \sqrt{x^2y\cdot\frac{1}{y}}\]
\[\Rightarrow x^2y+\frac{1}{y}\geq2x\]

Try reading this amazing note :