show $a^2 + b^2 + c^2 = 12 \implies a\sqrt[3]{b^2 + c^2} + b\sqrt[3]{c^2 + a^2} + c\sqrt[3]{a^2 + b^2} \le 12$

Question

Show that if $(a, b, c) \in \mathbf{R}_{\ge 0}$ such that $a^2 + b^2 + c^2 = 12$, the following inequality holds: $a\sqrt[3]{b^2 + c^2} + b\sqrt[3]{c^2 + a^2} + c\sqrt[3]{a^2 + b^2} \le 12$.

I tried using both Hölder, Cauchy, Power-Mean etc. but couldn't make much progress. I used power mean inequality in the following way:

\begin{equation} \left(\sum_{\text{cyc}} \frac{1}{3} \cdot a\sqrt[3]{b^2 + c^2}\right)^3 \le \frac{\sum_{\text{cyc}} a^3(b^2 + c^2)}{3} \end{equation}

So it would suffice to show that $\sqrt[3]{\frac{1}{3} \cdot \sum_{\text{cyc}} a^3(b^2 + c^2)} \le 4 \iff \sum_{\text{cyc}} a^3(b^2 + c^2) \le 192$, but from here I couldn't make much progress. I think the problem is that I never managed to use the fact $a^2 + b^2 + c^2 = 12$ to homogenize the inequality... I made another attempt using hölder inequality, but once again failed to homogenize... Here is that attempt:

\begin{equation} \left(\sum_{\text{cyc}} a^2\right)^{\frac{3}{2}} \cdot\left(\sum_{\text{cyc}} (b^2 + c^2)^{\frac{2}{3}} \right)^{\frac{3}{2}} \ge \left(\sum_{\text{cyc}} a\sqrt[3]{b^2 + c^2} \right)^3 \end{equation}

So we want to show that

\begin{equation} 12 = \sum_{\text{cyc}} a^2 \le \frac{\left(\sum_{\text{cyc}} a\sqrt[3]{b^2 + c^2} \right)^2}{\left(\sum_{\text{cyc}} (b^2 + c^2)^{\frac{2}{3}} \right)} \end{equation}

But here I'm stuck again... :(

Any help would be much appreciated :)

P.S the problem is from an olympiad handout about inequalities: https://web.evanchen.cc/handouts/Ineq/en.pdf

Answer 1

Let $a=2x$, $b=2y$ and $c=2z$.

Thus, $x^2+y^2+z^2=3$ and we need to prove that: $$\sum_{cyc}x\sqrt[3]{4(3-x^2)}\leq6$$ or $$\sum_{cyc}\left(2-x\sqrt[3]{4(3-x^2)}+\frac{2}{3}(x^2-1)\right)\geq0$$ or $$\sum_{cyc}\left(4+2x^2-3x\sqrt[3]{4(3-x^2)}\right)\geq0,$$ which is true by AM-GM: $$4+2x^2=2(2+x^2)=2\left(\frac{3-x^2}{2}+\frac{3x^2+1}{2}\right)\geq$$ $$\geq2\left(\frac{3-x^2}{2}+\frac{2x^2+2x}{2}\right)=2\left(\frac{3-x^2}{2}+x^2+x\right)\geq$$ $$\geq2\left(3\sqrt[3]{\frac{3-x^2}{2}\cdot x^2\cdot x}\right)=3x\sqrt[3]{4(3-x^2)}.$$

Answer 2

Some time since I asked this question, but looking back at it I found a very simple solution by Jensen, which I will leave up for anyone interested:

\begin{equation} \sum_{\text{cyc}} a\sqrt[3]{(b^2 + c^2)} \le 12 \iff \sum_{\text{cyc}}a\sqrt[3]{(12-a^2)} \le 12 \end{equation}

Now consider $f(x) = \sqrt{x} \cdot \sqrt[3]{(12-x)}$, which is concave on $[0, 12]$. Hence, Jensen gives (since $a^2, b^2, c^2 \in [0, 12]$)

\begin{equation} \frac{1}{3}\sum_{\text{cyc}}a\sqrt[3]{(12-a^2)} = \frac{f(a^2) + f(b^2) + f(c^2)}{3} \le f\left(\frac{a^2 + b^2 + c^2}{3}\right) = f(4) = 4 \\ \iff \sum_{\text{cyc}}a\sqrt[3]{(12-a^2)} = \sum_{\text{cyc}}a\sqrt[3]{(b^2+c^2)} \le 12 \end{equation} $\square$