Demetres wrote: Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=3$. Prove that \[ xyz(x+y+z)+2021\geqslant 2024xyz\] The following inequality a bit of stronger. Let $x$, $y$ and $z$ be non-negative numbers such that $x^2+y^2+z^2=3$. Prove that: $$4017xyz(x+y+z)+2021\geq14072xyz.$$

Let $p=x+y+z$ $q=xy+yz+zx$ and $r= xyz$. Fix $p$ and then $q$ becomes fixed. Then given inequality will be linear expression depended on $r$ . Then $max$ and $min$ values of equation will be obtained when $2$ of $x,y,z$ are equal. WLOG $y=z$. Then $x^2 + 2y^2 = 3$ it gives that $xy^2\leq 1$. Let's write the inequality as $x^2y^2+xy^3+xy^3+1+1+...+1\geq2024$$\sqrt[2021]{x^4y^8}$ ($2021$ times $1$). Then we need to prove that $\sqrt[2021]{x^4y^8}$$\geq$$xy^2$ Which is obvious, because $xy^2\leq 1$.

Demetres wrote: Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=3$. Prove that \[ xyz(x+y+z)+2021\geqslant 2024xyz\]

By AM-GM, we have \[x^2+y^2+z^2=3\geqslant 3\sqrt[3]{x^2y^2z^2} \implies xyz\leqslant 1.\]Let $w=\sqrt[3]{xyz}$, $0<w\leqslant 1$, \[xyz(x+y+z)+2021\geqslant 2024xyz \iff x+y+z\geqslant 2024-\frac{2021}{xyz},\]applying AM-GM, it's suffice to show that \[3w\geqslant 2024-\frac{2021}{w^3}\]which is \[(w-1)(3w^3-2021w-2021)\geqslant 0.\]Since $3w^3-2021w-2021< 3\cdot 1 -2021 \cdot 0-2021<0$ the inequality holds and equality occurs at $x=y=z=1$. $ \blacksquare$

We have $xyz\leq \left (\sqrt{\frac{x^2+y^2+z^2}{3}}\right )^3=1$ and $x+y+z \geq 3\sqrt[3]{xyz}$ by AM-GM. Let $xyz=a^3$, then we will prove $3a^4+2021-2024a^3 \geq 0$. By AM-GM, we have $a^4+a^4+a^4+1 \geq 4\sqrt[4]{a^{12}}=4a^3$. Then, $3a^4+2021-2024a^3 \geq 4a^3+2020-2024a^3=2020-2020a^3 \geq 0$ and we are done.

Generalization: Let $x,y,z,c$ be positive real numbers such that $x^2+y^2+z^2=3$. Prove that: $$cxyz(x+y+z)+2021 \ge (3c+2021)xyz$$

I think this is true too : Let $x,y,z,c$ be positive real numbers such that $x^2+y^2+z^2=3$. Prove that: $$c(x+y+z)+2021 \ge (3c+2021)xyz$$