# Prove that $a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$

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.

## 2 Answers

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

By AM-GM and Rearrangement we have:

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

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

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

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

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