prove that $3(a+b+c) \geq 8(a b c)^{1 / 3}+\left(\frac{a^{3}+b^{3}+c^{3}}{3}\right)^{1 / 3}$

Question

Question -

Suppose a,b,c are positive real numbers , prove that

$3(a+b+c) \geq 8(a b c)^{1 / 3}+\left(\frac{a^{3}+b^{3}+c^{3}}{3}\right)^{1 / 3}$

(Thailand $2006$)

My attempt -

we can assume that $a+b+c=1$ so we have to prove that

$3 \geq 8(a b c)^{1 / 3}+\left(\frac{a^{3}+b^{3}+c^{3}}{3}\right)^{1 / 3}$ but i am not able to show that it is true..

then i tried some AM-GM on RHS but none of them work, i think this is most different inequalities i have came across so i did not know where to go .

any help will be appreciated

thankyou

Answer 1

Also, we can use the following Holder: $$3(a+b+c)\geq\sqrt[3]{9^2(8abc+\frac{a^3+b^3+c^3}{3}})\geq8\sqrt[3]{abc}+\sqrt[3]{\frac{a^3+b^3+c^3}{3}},$$ where the first inequality it's just Muirhead or AM-GM.

Answer 2

Let $a+n+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $f(w^3)\leq0,$ where $$f(w^3)=8w+\sqrt{9u^3-9uv^2+w^3}-9u.$$ But, we see that $f$ increases, which says that it's enough to prove our inequality for a maximal value of $w^3$, which happens for equality case of two variables.

Since, our inequality is homogeneous, we can assume that $b=c=1$

because the case $b=c=0$ is trivial.

Can you end it now?

Answer 3

You don't need to assume a+b+c=1, you can prove this for genral case by applying am,gm inequality on a,b,c and a^3,b^3,c^3.