if $a^2+b^3+c^4+2019\ge k(a+b+c)$ find the maximum of the $k$

by geromty   Last Updated June 13, 2019 12:20 PM

Let $k$ be postive integers,and for any postive real number $a,b,c$ such $$a^2+b^3+c^4+2019\ge k(a+b+c)$$

find the maximum of $k$.

It seem use AM-GM inequality to solve it, but I can't find when is $=?$. So I can't solve it. Can you help.

Or is there any general way to find out about this nonhomogeneous inequality? Lagrangian multiplication does not seem to be able to handle this problem because the obtained derivative equation is difficult to solve: let $$f(a,b,c)=a^2+b^3+c^4+2019-k(a+b+c)$$ then $$f'_{a}=2a-k=0$$ $$f'_{b}=3b^2-k=0$$ $$f'_{c}=4c^3-k=0$$ so we have $$2a=3b^2=4c^3$$ then which $a,b,c$ can find the maximum of the $k?$ Thanks



Answers 1


For $(a,b,c)=(40,5,3)$ we obtain $k<80.$

We'll prove that $k=79$ is valid, for which we need to prove that $$a^2+b^3+c^4+2019\geq79(a+b+c)$$ or $$(a^2-79a+1561)+(b^3-79b+271)+(c^4-79c+187)\geq0,$$ which is true by AM-GM.

Can you end it now?

Michael Rozenberg
Michael Rozenberg
June 13, 2019 12:17 PM

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