Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that $$a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$$ I found a point at which the equality is attended, say $a=0,b=1,c=2$. But I have no idea how to prove it. I tried to use the AM-GM inequality but then I obtained the more difficult one. Please help me. Thank you very much.


$\text{WLOG b=mid(a,b,c)} $

By AM-GM and Rearrangement we have:

$$\text{L.H.S}=\sum _{cyc}a\sqrt{b^3+1}=\sum _{cyc}a\sqrt{\left(b+1\right)\left(b^2-b+1\right)}$$

$$\le \sum _{cyc}a\cdot \frac{b+1+b^2-b+1}{2}=\sum_{cyc}\frac{2a+ab^2}{2}=\frac{ab^2+bc^2+ca^2}{2}+3$$

$$\le \frac{b\left(a^2+ac+c^2\right)}{2}+3\le \frac{b\left(a+c\right)^2}{2}+3$$

$$=\frac{2b\left(a+c\right)^2}{4}+3\le \frac{\left(\frac{2\left(a+b+c\right)}{3}\right)^3}{4}+3=5=\text{R.H.S}$$

  • $\begingroup$ In the middle, you use $ab^2 + ca^2 \le ba^2 + abc$. Can you explain? $\endgroup$ – Andreas Mar 18 at 16:57
  • $\begingroup$ $ab^2+bc^2+ca^2\le a^2b +abc+bc^2$ It is rearrangement inequality. See here: en.wikipedia.org/wiki/Rearrangement_inequality and here brilliant.org/wiki/rearrangement-inequality $\endgroup$ – Word Shallow Mar 18 at 16:59
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    $\begingroup$ Yes, takes a little reordering, I got it. $\endgroup$ – Andreas Mar 18 at 17:02
  • $\begingroup$ I'm sorry, but I think you can not assume that $a\geq b\geq c$ since the inequality is not symmetric. $\endgroup$ – Maria Mazur Mar 18 at 20:19
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    $\begingroup$ Here we can assume $b$ is middle among $a,b,c$ and get the result. Thank you. $\endgroup$ – RuaSun Mar 18 at 22:47

Start out as before: $$ a\sqrt{b^{3}+1}+b\sqrt{c^{3}+1}+c\sqrt{a^{3}+1}\leq5 $$

$$ a\cdot\sqrt{\left(b+1\right)\left(b^{2}-b+1\right)}\leq a\cdot\frac{b^{2}+2}{2} $$

$$ \frac{a\left(b^{2}+1\right)+b\left(c^{2}+1\right)+c\left(a^{2}+1\right)}{2}=3+\frac{ab^{2}+bc^{2}+ca^{2}}{2} $$ Continue by lagrange multipliers: $$ \Lambda=ab^{2}+bc^{2}+ca^{2}+\lambda\left(a+b+c-3\right) $$ $$ \begin{cases} \partial_{a}\Lambda=b^{2}+2ac-\lambda & \Rightarrow\lambda=b^{2}+2ac\\ \partial_{b}\Lambda=c^{2}+2ab-\lambda & \Rightarrow\lambda=c^{2}+2ab\\ \partial_{c}\Lambda=a^{2}+2bc-\lambda & \Rightarrow\lambda=a^{2}+2bc \end{cases} $$

$$ H=\left(\begin{array}{cccc} 2c & 2b & 2a & 1\\ 2b & 2a & 2c & 1\\ 2a & 2c & 2b & 1\\ 1 & 1 & 1 & 0 \end{array}\right) $$

$$ b^{2}+2ac=c^{2}+2ab\Rightarrow(b-c)\left(b+c-2a\right)=0 $$

$$ (b-c)\left(b+c-2a\right)=(a-c)\left(a+c-2b\right)=(a-b)\left(a+b-2c\right)=0 $$

If we have no two equal, then we arrive at a contradiction:

$$ b+c-2a=a+c-2b=a+b-2c=0\Rightarrow a=b=c $$

Hence assume $a-b=0$, then either $a-c=0$ and getting $a=b=c=1$, or: $a+c-2a=0\Rightarrow a=c$ yielding the same results. Hence the only candidate where $abc\neq0$ is:

$$ a=b=c=1 $$

Here, we get:

$$ ab^{2}+bc^{2}+ca^{2}=3 $$

Assume $b=0$ if we are on the boundary, then by AM-GM:

$$ ab^{2}+bc^{2}+ca^{2}=ca^{2}=\frac{1}{2}\left(2\left(3-a\right)a^{2}\right)\leq\frac{1}{2}\left(\frac{2\left(3-a\right)+a+a}{3}\right)^{3}=4 $$

Hence we have

$$ \frac{ab^{2}+bc^{2}+ca^{2}}{2}\leq2 $$

As needed (maybe not too detailed why the $a=b=c=1$ is a saddle point). THe surface is shown below: Plotted out the surface in 2 variables


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