Let $a$, $b$ and $c$ be positive real numbers.

$(i)$ Prove that $4(a^3 + b^3) \geq (a+b)^3$

$(ii)$ Prove that $9(a^3 + b^3 + c^3) \geq (a+b+c)^3$

## Solution

For $(i)$, we have

\begin{align} a^3+b^3 &\geq& a^3+b^3 \label{y1996r1p5e1}\\ 3(a^3+b^3) &\geq& 3(ab^2 + ba^2) \label{y1996r1p5e2} \end{align}

$\ref{y1996r1p5e1}$ is trivial.

We get $\ref{y1996r1p5e2}$ from Muirhead’s inequality because $(3,0)$ majorizes $(2,1)$.

Adding $\ref{y1996r1p5e1}$ and $\ref{y1996r1p5e2}$, we get $(i)$.

For $(ii)$, we have \begin{align} a^3+b^3+c^3 &\geq& a^3+b^3+c^3 \label{y1996r1p5e3}\\ 6(a^3+b^3+c^3) &\geq& 3(a^2b + a^2c+b^2a+b^2c+c^2a+c^2b) \label{y1996r1p5e4}\\ 2(a^3+b^3+c^3) &\geq& 6(abc) \label{y1996r1p5e5} \end{align}

$\ref{y1996r1p5e3}$ is trivial.

We get $\ref{y1996r1p5e4}$ from Muirhead’s inequality because $(3,0)$ majorizes $(2,1)$.

We have $\ref{y1996r1p5e5}$, because $AM \geq GM$.

Adding $\ref{y1996r1p5e3}$,$\ref{y1996r1p5e4}$ and $\ref{y1996r1p5e5}$, we get $(ii)$.