Maximum value problem involving three variable

Question

I have faced this problem

Let $a, b, c$ be three positive real numbers satisfy $a^2+b^2+c^2=k$ ($k$ is a constant). Find the maximum value of $S=\sqrt{1+a^3}+\sqrt{1+b^3}+\sqrt{1+c^3}$.

I found a solution as follows

Let $f(x)=\sqrt{1+x^3}$ then the second derivative of $f$: $f''(x)=\dfrac{-3x^2}{4\sqrt{(1+x^3)^3}}<0\ \forall x>0 $.

Thus $f(x)$ is a concave function.

By Jensen inequality we have $f(a)+f(b)+f(c)\le 3f\left(\dfrac{a+b+c}{3}\right)$ then $$ \sqrt{1+a^3}+\sqrt{1+b^3}+\sqrt{1+c^3} \le 3\sqrt{1+\left(\dfrac{a+b+c}{3}\right)^3}$$ By Cauchy-Schwarz inequality we have $a+b+c\le \sqrt{3(a^2+b^2+c^2)}= \sqrt{3k}$ then $$ \sqrt{1+a^3}+\sqrt{1+b^3}+\sqrt{1+c^3} \le 3\sqrt{1+\left(\dfrac{\sqrt{3k}}{3}\right)^3} = 3\sqrt{1+\sqrt{\dfrac{k^3}{27}}} $$ Equality occurs if and only if a = b = c.

Thus the maximum value of $S$ is $3\sqrt{1+\sqrt{\dfrac{k^3}{27}}}$ at $a=b=c=\sqrt{\dfrac{k}{3}}$.

My questions are

1. Is my solution correct?
2. Is there any more elementary solution with AM-GM; Cauchy-Schwarz; ... which doesn't involve derivative, concave function, Jensen inequality etc?

My attempt for the 2nd question, I have found out for the special case $k=12$ when I tried to find $x, y$ such that $x\sqrt{1+a^3}\le a^2 + y$:

By AM-GM inequality, we have $a^2+4\ge 4a$, then $a^4+4a^2\ge 4a^3$ and therefore $$ a^4+4a^2+4\ge 4(1+a^3)\Leftrightarrow (a^2+2)^2\ge 4(1+a^3)\Leftrightarrow a^2+2 \ge 2\sqrt{1+a^3}$$

Similarly, $b^2+2 \ge 2\sqrt{1+b^3}; c^2+2 \ge 2\sqrt{1+c^3}$, hence $$ \sqrt{1+a^3}+\sqrt{1+b^3}+\sqrt{1+c^3} \le \dfrac{1}{2}\left( 6 +a^2+b^2+c^2 \right) = 9 $$ Equality occurs if and only if $a=b=c=2$ then the maximum value of $S$ is 9.

Please help me. Thanks

Answer 1

Let $a^2=x$, $b^2=y$, $c^2=z$ and $f(x)=\sqrt{x^{\frac{3}{2}}+1}.$

Thus, $x+y+z=k$ and we need to find $\max\sum\limits_{cyc}f(x).$

But $$f''(x)=\frac{3\left(2-\sqrt{x^3}\right)}{16\sqrt{x}\sqrt{\left(\sqrt{x^3}+1\right)^3}}.$$

1. $k\leq\sqrt[3]4.$

In this case $f''(x)\geq0,$ which says $f$ is a convex function.

Let $x\geq y\geq z$.

Thus, since $(k,0,0)\succ(x,y,z)$, by Karamata we obtain: $$\sum_{cyc}\sqrt{a^3+1}=\sum_{cyc}f(x)\leq f(k)+2f(0)=\sqrt{\sqrt{k^3}+1}+2.$$ 2. $k>\sqrt[3]4$.

In this case on $[0,k]$ the function $f$ has an unique inflaction point,

which by the Vasc's HCF Theorem says that $f$ gets a maximal value for equality case of two variables.

Can you end it now?