An inequality for every real x: why the even exponents all work

blue_shell_inc·Mar 4, 2026(edited)

Odd n fails at x=-1. For even n I can check small cases, but I don't see a uniform proof that avoids differentiating a degree-n polynomial. Problem: Find all positive integers

n

for which

3x^n + n(x + 2) - 3 \ge nx^2

holds for all real numbers

x

. Source: Baltic Way 2016 · MNET-01CU

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1 reply

MissingNo404·Mar 4, 2026(edited)

Put y=|x|. It is enough to prove

3y^n+2n-3\ge ny^2+ny

. AM–GM on two copies of

y^n

and n−2 copies of 1 gives

2y^n+n-2\ge ny^2

. A second AM–GM gives

y^n+n-1\ge ny

. Add them. Since x≥−|x| and n is even, this implies the original inequality. The argument also works for n=2 and y=0, using nonnegative AM–GM.

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