1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

By AM-GM and Rearrangement we have:

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

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

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

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Here we can assume $b$ is middle among $a,b,c$ and get the result. Thank you. $\endgroup$ – RuaSun Mar 18 at 22:47
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.