Minimum of $abc$ when $a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}.$

Let $$a,b,c$$ be positive real numbers with $$abc=k$$ such that the inequality $$a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$$ holds for all $$a,b,c$$. Find the minimum value of $$k$$.

I found that $$abc=2$$ works. Here is my proof for $$abc=2$$:

By Cauchy-Schwarz and Muirhead inequality \begin{align} a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b} &\leq \sqrt{2(a^2+b^2+c^2)(a+b+c)}\\ &= \sqrt{abc(a^2+b^2+c^2)(a+b+c)}\\ &= \sqrt{\sum_{cyc}a^4bc+\sum_{cyc}(a^3b^2c+a^2b^3c)}\\ &\leq \sqrt{\sum_{cyc}a^6+2\sum_{cyc}a^3b^3}\\ &= a^3+b^3+c^3 \end{align} as desired. Equality holds when $$a=b=c=\sqrt[3]2$$.

The inequality seems quite tight also. But I am not sure if it is the minimum of $$abc$$. And if it is, how do I prove that?

• Looks like it, for $a=b=c$, the inquality becomes $k \ge \sqrt{2k}$. Sep 29, 2021 at 18:36
• There is something wrong with your proof. You seem to get that for any $a,b,c$ the inequality holds. Let's choose $a=b=c=1$. Then you get $3\ge 3\sqrt 2$. This is false, Sep 29, 2021 at 19:46
• @Andrei In the second line of the inequality, I substituted $2$ with $abc$. So the inequality holds only when $abc=2$ not for any $a,b,c$. Sep 30, 2021 at 4:55

Let $$abc=k$$.
Thus, after homogenization we need to prove that $$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{k}}\right)\geq0,$$ which for $$a=b=c$$ gives $$k\geq2$$ and it's enough to prove that: $$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{2}}\right)\geq0,$$ which you proved already by C-S.
Also, it's true by AM-GM and Muirhead: $$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{2}}\right)=\sum_{cyc}\left(a^3-a\sqrt{\frac{2bc(ab+ac)}{4}}\right)\geq$$ $$\geq\sum_{cyc}\left(a^3-\frac{a(2bc+ab+ac)}{4}\right)=\frac{1}{4}\sum_{cyc}(4a^3-a^2b-a^2c-2abc)\geq0.$$