If $a=b=c=d=-1$ then $-4\geq -4k$ $\Rightarrow$ $k\geq 1$. If $a=b=c=d=1/2$ then $1/2 \geq 4k$ $\Rightarrow$ $k\leq 1/8$. So there are no such $k$. What am I missing?

I know!!! Valentin wrote wrong problem text. Right text: Find all reals $k$ such that \[a^3+b^3+c^3+d^3+1\geq k(a+b+c+d)\] holds for all $a,b,c,d\geq -1$.

According to the Chinese version, myth's correction is right.

Putting a=b=c=d=-1 gives k>=3/4 Putting a=b=c=d=1/2 gives 3/4>=k.So k=3/4. Also for k=3/4 the inequality becomes (2a-1) ² (a+1)+(2b-1) ² (b+1)+(2c-1) ²(c+1)+(2d-1) ²(d+1)>=0 which is quite true.

Many contestants in the competition just proved if such $k$ exists, then it will be $3\over4$, and thus losing many points. Luckily, I didn't make this mistake!

The better idea is to put $k=\sqrt{abcd}$ [from Muirhead] and it will be probably easier to find all $k$ am I wrong ?

Yes, you are wrong, because $a,b,c,d$ can be negative.

... this was evidence of my stupidity

"Before you attempt a problem, think about it" -- someone, probably Notice that it suffices to choose $k$ such that the inequality $$a^3+\frac 14 \geq ka$$for all $a \geq -1$. In other words, it suffices to find $k$ such that $$a^2 + \frac 1{4a} \geq k$$for $a \geq 0$, and the reverse inequality sign otherwise. Indeed, the two inequalities yield $k \leq \frac 34$ and $k \geq \frac 34$ respectively, so we must have $k = \frac 34$.

Valentin Vornicu wrote: Find all reals $ k$ such that\[ a^3 + b^3 + c^3 + d^3 + 1\geq k(a + b + c + d)\]holds for all $ a,b,c,d\geq - 1$. When $a=b=c=d=-1,$ $-3=a^3 + b^3 + c^3 + d^3 + 1\geq k(a + b + c + d)=-4k\implies k\geq \frac{3}{4}$ When $a=b=c=d=\frac{1}{2},$ $\frac{3}{2}=a^3 + b^3 + c^3 + d^3 + 1\geq k(a + b + c + d)=2k\implies k\leq \frac{3}{4}$ $$k=\frac{3}{4}.$$$$(a+1)(2a-1)^2\geq 0\iff a^3+\frac{1}{4}\geq \frac{3}{4} $$$$a^3 + b^3 + c^3 + d^3 + 1\geq \frac{3}{4}(a + b + c + d)$$Equality holds when $a=b=c=d=\frac{1}{2}.$

Let $ a,b,c,d\geq - 1,$ Prove that$$a^3 + b^3 + c^3 + d^3+8\geq (a+b)(c+ d)$$$$a^3 + b^3 + c^3 + d^3+\frac{25}{4}\geq (a+b+\frac{1}{2})(c+ d+\frac{1}{2})$$

sqing wrote: Let $ a,b,c,d\geq - 1,$ Prove that$$a^3 + b^3 + c^3 + d^3+8\geq (a+b)(c+ d)$$ $$\left((a-1)^2+1\right)(a+1) \geq 0 \iff a^3+2\geq a^2$$\begin{align*}\Longrightarrow a^3+b^3+c^3+d^3+8 & =(a^3+2)+(b^3+2)+(c^3+2)+(d^3+2) \\ &\geq a^2+b^2+c^2+d^2 \\ &= (a+b)(c+d)+\frac{1}{2}\left((a-c)^2+(b-c)^2+(a-d)^2+(b-d)^2\right) \\ &\geq (a+c)(b+d)\end{align*} sqing wrote: $$a^3 + b^3 + c^3 + d^3+\frac{25}{4}\geq (a+b+\frac{1}{2})(c+ d+\frac{1}{2})$$ $$\left((a-1)^2+\frac{1}{2}\right)(a+1)\geq 0 \iff a^3+\frac{3}{2} \geq a^2+\frac{1}{2}a$$\begin{align*} \Longrightarrow a^3+b^3+c^3+d^3+\frac{25}{4} &= (a^3+\frac{3}{2})+(b^3+\frac{3}{2})+(c^3+\frac{3}{2})+(d^3+\frac{3}{2})+\frac{1}{4}\\ &\geq a^2+\frac{1}{2}a+b^2+\frac{1}{2}b+c^2+\frac{1}{2}c+d^2+\frac{1}{2}d+\frac{1}{4} \\ &= (a+b)(c+d)+\frac{1}{2}(a+b)+\frac{1}{2}(c+d)+\frac{1}{4}+\frac{1}{2}\left((a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2\right) \\& \geq (a+b)(c+d)+\frac{1}{2}(a+b)+\frac{1}{2}(c+d)+\frac{1}{4} \\ & = (a+b+\frac{1}{2})(c+d+\frac{1}{2}) \end{align*}

Valentin Vornicu wrote: Find all reals $ k$ such that \[ a^3 + b^3 + c^3 + d^3 + 1\geq k(a + b + c + d) \]holds for all $ a,b,c,d\geq - 1$. Edited by orl. Nice one. I also provided a proof, because it was reposted one year ago

Let $ a,b,c,d\geq - 1,$ Prove that$$a^3 + b^3 + c^3 + d^3+10\geq (a+2b)(c+d)$$$$a^3 + b^3 + c^3 + d^3+14\geq (3a+2b)(c+d)$$$$a^3 + b^3 + c^3 + d^3+\frac{141}{16}\geq (a+2b+\frac{1}{4})(c+ d+\frac{1}{4})$$$$a^3 + b^3 + c^3 + d^3+\frac{169}{16}\geq (2a+2b+\frac{1}{4})(c+ d+\frac{1}{4})$$

HamstPan38825 wrote: "Before you attempt a problem, think about it" -- someone, probably Notice that it suffices to choose $k$ such that the inequality $$a^3+\frac 14 \geq ka$$for all $a \geq -1$. In other words, it suffices to find $k$ such that $$a^2 + \frac 1{4a} \geq k$$for $a \geq 0$, and the reverse inequality sign otherwise. Indeed, the two inequalities yield $k \leq \frac 34$ and $k \geq \frac 34$ respectively, so we must have $k = \frac 34$. Nice

Let $ a,b,c,d\geq - 1,$ Prove that$$a^3 + b^3 + c^3 + d^3+6\geq \frac{1}{2}(a+b)(c+ d)$$$$a^3 + b^3 + c^3 + d^3+\frac{113}{16}\geq (a+b+\frac{1}{4})(c+ d+\frac{1}{4})$$