Proving an inequality with convexity

Question

Hello i struggle prooving this inequality

$$abc\leq\frac13 (a^3+b^3+c^3)$$ $(a,b,c)$ are positive reals I tought about using the fact that the exponential is convex and the jersen inequality but i have some problems and i couldn't prove this. Maybe using the cube function instead may work on $$\mathbb {R}^{+*} $$ Thanks for the help

Answer 1

By concavity of $\ln$ on $(0, +\infty)$, you have $$\ln \left( \frac{1}{3} \left( a^3 + b^3 + c^3 \right)\right) \geq \frac{1}{3} \left(\ln(a^3) +\ln(b^3)+ \ln(c^3)\right) = \ln(abc)$$

Now take the exponential and you are done.

Answer 2

Apply the AM-GM inequality to $a^3$, $b^3$, $c^3$.

Answer 3

Also, we can use Jensen for the convex function $f(x)=e^x$.

Indeed, let $a=e^{\frac{x}{3}},$ $b=e^{\frac{y}{3}}$ and $c=e^{\frac{z}{3}}.$

Thus, by Jensen we obtain: $$\frac{a^3+b^3+c^3}{3}=\frac{e^x+e^y+e^z}{3}=\frac{f(x)+f(y)+f(z)}{3}\geq$$ $$\geq f\left(\frac{x+y+z}{3}\right)=e^{\frac{x+y+z}{3}}=e^{\frac{x}{3}}e^{\frac{y}{3}}e^{\frac{z}{3}}=abc.$$