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There's AM-GM inequality. But I've bumped into AM-CM inequality in official solutions for IMO 2018 problems. Here:

enter image description 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?

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    $\begingroup$ CM probably stands for cubic mean. $\endgroup$ – Qi Zhu Jul 27 at 15:59
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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.

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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".

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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 $a=\sqrt[3]{\frac{x+t+14}{7}}$ and $b=\sqrt[3]{\frac{y+t+14}{7}}$.

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  • $\begingroup$ Thanks for the REAL Shortlist! :) $\endgroup$ – user75619 Jul 27 at 20:16
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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 $\varphi(x) = \sqrt[3]{x}$.

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