Prove that for all positive reals $a, b,c$ we have: $a +\sqrt{ab}+ \sqrt[3]{abc}\le \frac43 (a + b + c)$
Problem
Source: Dutch IMO TST 2016 day 2 p1
Tags: algebra, inequalities
30.08.2019 14:03
parmenides51 wrote: Prove that for all positive reals $a, b,c$ we have: $a +\sqrt{ab}+ 3\sqrt[3]{abc}\le \frac43 (a + b + c)$ There is a typo i guess? $(1,1,1)$ is a clear contradiction.
30.08.2019 14:08
yes you are correct, the 3 factor is wrong, shall edit it 1st post with the typo parmenides51 wrote: Prove that for all positive reals $a, b,c$ we have: $a +\sqrt{ab}+ 3\sqrt[3]{abc}\le \frac43 (a + b + c)$
30.08.2019 14:08
parmenides51 wrote: Prove that for all positive reals $a, b,c$ we have: $a +\sqrt{ab}+ \sqrt[3]{abc}\le \frac43 (a + b + c)$ Slovenia 2012
31.08.2023 22:35
You can also finish this with Kiran Kedlaya's result and assuming that $c\ge b\ge a$
13.04.2024 13:17
parmenides51 wrote: Prove that for all positive reals $a, b,c$ we have: $a +\sqrt{ab}+ \sqrt[3]{abc}\le \frac43 (a + b + c)$ By AM-GM, we have \begin{align*} a &\le a \\ \sqrt{ab} &\le \frac{1}{4}a + b \\ \sqrt[3]{abc} &\le \frac{1}{12}a + \frac{1}{3}b + \frac{4}{3}c \end{align*}add these three inequalities to obtain the desired result $a +\sqrt{ab}+ \sqrt[3]{abc}\le \frac43 (a + b + c)$
02.08.2024 00:12
Let $b=ax^2, c=axy^3$ then we need to prove $0\leq 4xy^3+4x^2+1-3xy-3x=(2x-1)^2+x(y+1)(2y-1)^2$, and we're done!
20.09.2024 10:21
Adding the following inequalities due to AM-GM, $$a = a$$$$\sqrt{ab} \leq \frac{1}{2}(\frac{1}{2}+2b)$$$$\sqrt[3]{abc} \leq \frac{1}{3} (\frac{1}{4}a+b+4c)$$we get the desired inequality.