6

Let $a,b,c\in\mathbf R^+$, such that $a+b+c=3$. Prove that $$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$
$Hint$ : Use Rearrangement Inequality

My Work :-$\\$
Without Loss of Generality let's assume that $0\le a\le b\le c$ then we can infer that $c^2\ge b^2\ge a^2$ also $b+c\ge c+a \ge a+b$. Thus $\frac{1}{b+c}\le \frac{1}{c+a}\le \frac{1}{a+b}$. Hence by Rearrangement Inequality $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}$ is the greatest permutation however I am not able to express $\frac{a^2+b^2+c^2}{2}$ as another permutation of the same, hence I require assistance.

Thank You

Martin Sleziak
  • 56,568
arnav_de
  • 719

4 Answers4

9

I'm not sure how to use Rearrangement Inequality but I can do it with tangent line method.

Using $a+b+c=3$ we can rewrite OP as: $$\underbrace{{a^2\over 3-a}-{a^2\over 2}}_{f(a)} + {b^2\over 3-b}- {b^2\over 2}+{c^2\over 3-c} -{c^2\over 2} \geq 0$$ If you calculate a tangent at point $x=1$ on $f(x)$ we get $y={1\over 4} (x-1)$ and it is not hard to see that for $x\in (0,3)$ we have $$f(x)\geq {1\over 4}(x-1)$$

Using this we get $$f(a) +f(b)+f(c)\geq {1\over 4}(a-1)+ {1\over 4}(b-1)+{1\over 4}(c-1)=0$$

nonuser
  • 92,607
6

Since triples $(a^2,b^2,c^2)$ and $\left(\frac{1}{b+c},\frac{1}{a+c},\frac{1}{a+b}\right)$ have the same ordering,

by Rearrangement and C-S we obtain: $$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{1}{3}\sum_{cyc}a^2\sum_{cyc}\frac{1}{b+c}\geq\frac{1}{3}\sum_{cyc}a^2\cdot\frac{(1+1+1)^2}{\sum\limits_{cyc}(b+c)}=\frac{a^2+b^2+c^2}{2}.$$

Michael Rozenberg
  • 1
  • The first inequality you use isn't it called chebyshev inequality rather than rearrangement? I understand that they are very similar though... – Thomas Jul 23 '21 at 09:04
  • 1
    The Chebyshov's inequality follows immediately from Rearrangement. But you are right, it's indeed Chebyshov – Michael Rozenberg Jul 23 '21 at 09:43
5

Another way:

We need to prove that: $$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{3(a^2+b^2+c^2)}{2(a+b+c)}$$ or $$\sum_{cyc}\left(\frac{a^2}{b+c}-\frac{3a^2}{2(a+b+c)}\right)\geq0$$ or $$\sum_{cyc}\frac{a^2(a-b-(c-a))}{b+c}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{a^2}{b+c}-\frac{b^2}{a+c}\right)\geq0,$$ which is true because $(a,b)$ and $\left(\frac{a^2}{b+c},\frac{b^2}{a+c}\right)$ have the same ordering for any positive $c$.

Michael Rozenberg
  • 1
1

I apply rearrangement inequality 3 times to obtain

$$ 3\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right) \geqq (a^2+b^2+c^2)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) $$

Next we can show that

$$ \frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\geqq \frac{3}{2} $$

by using Cauchy-Schwarz inequality and the fact that $a+b+c=3$.

Combining both inequalities give the desired result.

Li Kwok Keung
  • 5,623