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


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?

The play with these numbers you can make by the following way.

You got that for $a=\frac{k}{2}$, $b=\sqrt{\frac{k}{3}}$ and $c=\sqrt[3]{\frac{k}{4}}$ we obtain a critical point.

In this point should be: $$\left(\frac{k}{2}\right)^2+\left(\sqrt{\frac{k}{3}}\right)^3+\left(\sqrt[3]{\frac{k}{4}}\right)^4+2019-k\left(\frac{k}{2}+\sqrt{\frac{k}{3}}+\sqrt[3]{\frac{k}{4}}\right)\geq0.$$ Now, we can see that for $k=79$ this inequality is still true, but for $k=80$ it's wrong.

Thus, for the critical point, when $k=79$ we obtain: $$(a,b,c)=\left(39.5, 5.13..., 2.70...\right)\approx(40,5,3)$$ and from here you can get a proof.

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