# What is AM-CM inequality?

There's AM-GM inequality. But I've bumped into AM-CM inequality in official solutions for IMO 2018 problems. Here: This is A7 problem, solution 1 from here: https://www.imo-official.org/problems/IMO2018SL.pdf

I attempted to use AM-GM inequality in case they simply misspelled, but AM-GM doesn't seem to work here.

Any ideas?

• CM probably stands for cubic mean. – Qi Zhu Jul 27 at 15:59

The AM-CM inequality likely denotes the Arithmetic mean - Cubic mean inequality which is a specific instance of the Generalized mean inequality. This inequality follows directly from Jensen's inequality mentioned in other answers.

Given what is going on, what they are using is the fact that if a function $$f:I\to\mathbb R$$ is convex (here $$I\subset\mathbb R$$ is an interval), and $$x,y\in I$$, then $$f\left(\frac{x+y}{2}\right)\leq\frac{f(x)+f(y)}{2}$$ or, equivalently, if $$f:I\to\mathbb R$$ is concave, then $$\frac{f(x)+f(y)}{2}\leq f\left(\frac{x+y}{2}\right).$$ Some people take the statement as the definition of a convex function, and then a function $$f$$ is concave if $$-f$$ is convex.

I imagine the "C" in AM-CM stands for either "convex" or "concave".

Elementary approach: no Jensen's inequality or AM-GM inequality (or even AM-CM inequality). BTW I warmly recommend the REAL Shortlist 2018.

It suffices to show that if $$a,b\geq 0$$ then $$\left(\frac{a+b}{2}\right)^3\leq \frac{a^3+b^3}{2}\tag{1}$$ which is equivalent to $$0\leq a^3+b^3-a^2b-ab^2=(a+b)(a-b)^2$$ that trivially holds.

Then apply (1) by letting $$a=\sqrt{\frac{x+t+14}{7}}$$ and $$b=\sqrt{\frac{y+t+14}{7}}$$.

• Thanks for the REAL Shortlist! :) – user75619 Jul 27 at 20:16

They appear to be using Jensen's inequality with (Wikipedia's notation) $$x_1 = \frac{x+t+14}{7}$$, $$x_2=\frac{y+z+14}{7}$$ and $$a_1 = a_2 = 1$$ and $$\varphi(x) = \sqrt{x}$$.