Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.

Question

$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}{a + b}$$

I have not come up with any ideas to solve the problem yet. I will probably in the near future but right now, I can't.

Answer 1

SOS helps!

Let $a=b=c=\frac{1}{3\sqrt2}.$

Thus, we get a value $\frac{1}{2\sqrt2}.$

We'll prove that it's a minimal value.

Indeed, we need to prove that $$\sum_{cyc}\frac{a^2}{b+c}\geq\frac{1}{2\sqrt2}\sum_{cyc}\sqrt{a^2+b^2}$$ or $$\sum_{cyc}\left(\frac{a^2}{b+c}-\frac{a}{2}\right)\geq\frac{1}{4}\sum_{cyc}\left(\sqrt{2(a^2+b^2)}-a-b\right)$$ or $$\sum_{cyc}\frac{a(a-b-(c-a))}{b+c}\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ or $$\sum_{cyc}(a-b)\left(\frac{a}{b+c}-\frac{b}{a+c}\right)\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ or $$\sum_{cyc}\frac{(a-b)^2(a+b+c)}{(a+c)(b+c)}\geq\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{\sqrt{2(a^2+b^2)}+a+b}$$ and since by C-S $$\sqrt{2(a^2+b^2)}=\sqrt{(1^2+1^2)(a^2+b^2)}\geq a+b,$$ it's enough to prove that: $$\sum_{cyc}\frac{(a-b)^2(a+b+c)}{(a+c)(b+c)}\geq\frac{1}{4}\sum_{cyc}\frac{(a-b)^2}{a+b}$$ or $$\sum_{cyc}(a-b)^2(4(a+b+c)(a+b)-(a+c)(b+c))\geq0,$$ for which it's enough to prove that $$\sum_{cyc}(a-b)^2(a^2+b^2-c^2)\geq0.$$ Now, let $a\geq b\geq c$.

Thus, $$\sum_{cyc}(a-b)^2(a^2+b^2-c^2)\geq(a-c)^2(a^2+c^2-b^2)+(b-c)^2(b^2+c^2-a^2)\geq$$ $$\geq(b-c)^2(a^2-b^2)+(b-c)^2(b^2-a^2)=0$$ and we are done!

Answer 2

I have an alternative solution, by CS:

$$A=\sum\limits_{cyc} \frac{a^2}{b+c}=\sum\limits_{cyc} \frac{a^4}{a^2b+a^2c}\geq \frac{\left(\sum\limits_{cyc}a^2\right)^2}{\sum\limits_{sym}a^2b} $$

By Jensen:

$$1=\sum\limits_{cyc}\sqrt{a^2+b^2}\leq 3\sqrt{\frac{2}{3}(a^2+b^2+c^2)}$$

So

$$A\geq \frac{\left(\sum\limits_{cyc}a^2\right)^2}{(\sum\limits_{sym}a^2b )* 1}\geq \frac{1}{\sqrt{6}}\frac{\left(\sum\limits_{cyc}a^2\right)^{\frac{3}{2}}}{\sum\limits_{sym}a^2b}$$

If you plug in $a=b=c=\frac{1}{3\sqrt{2}}$ you will get $A=\frac{1}{2\sqrt{2}}$

We will show that $\frac{1}{\sqrt{6}}\frac{\left(\sum\limits_{cyc}a^2\right)^{\frac{3}{2}}}{\sum\limits_{sym}a^2b}\geq \frac{1}{2\sqrt{2}}$

Or $$4\left(\sum\limits_{cyc}a^2\right)^{3}\geq 3\left(\sum\limits_{sym}a^2b\right)^2$$

Which is true after foiling by Muirhead and Schur

Answer 3

By rearrangement-inequality

$$\frac{a^{\,2}}{b+ c}+ \frac{b^{\,2}}{c+ a}+ \frac{c^{\,2}}{a+ b}\geqq \frac{a^{\,2}}{a+ b}+ \frac{b^{\,2}}{b+ c}+ \frac{c^{\,2}}{c+ a}$$

We have to prove

$$\frac{a^{\,2}}{a+ b}+ \frac{b^{\,2}}{b+ c}+ \frac{c^{\,2}}{c+ a}\geqq \frac{1}{2}(\,\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}+ \sqrt{\frac{b^{\,2}+ c^{\,2}}{2}}+ \sqrt{\frac{c^{\,2}+ a^{\,2}}{2}}\,)$$

By a.m.-g.m.-inequality

$$\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}= \frac{2\sqrt{2(\,a^{\,2}+ b^{\,2}\,)(\,a+ b\,)^{\,2}}}{4(\,a+ b\,)}\leqq \frac{(\,a+ b\,)^{\,2}+ 2(\,a^{\,2}+ b^{\,2}\,)}{4(\,a+ b\,)}$$

We need to have

$$\frac{4\,a^{\,2}}{a+ b}+ \frac{4\,b^{\,2}}{b+ c}+ \frac{4\,c^{\,2}}{c+ a}\geqq \frac{1}{2}\times \sum\limits_{cyc}\frac{(\,a+ b\,)^{\,2}+ 2(\,a^{\,2}+ b^{\,2}\,)}{(\,a+ b\,)}$$

$$\because\,\frac{a^{\,2}- 2\,ab+ b^{\,2}}{a+ b}+ \frac{b^{\,2}- 2\,bc+ c^{\,2}}{b+ c}+ \frac{c^{\,2}- 2\,ca+ a^{\,2}}{c+ a}\geqq 0$$

So it holds !

$$\therefore\,\frac{a^{\,2}}{b+ c}+ \frac{b^{\,2}}{c+ a}+ \frac{c^{\,2}}{a+ b}\geqq \frac{1}{2}(\,\sqrt{\frac{a^{\,2}+ b^{\,2}}{2}}+ \sqrt{\frac{b^{\,2}+ c^{\,2}}{2}}+ \sqrt{\frac{c^{\,2}+ a^{\,2}}{2}}\,)= \frac{1}{2\sqrt{2}}$$

Answer 4

By rearrangement inequality,

$$\sum_{cyc} \frac{a^2}{b+c} \ge \sum_{cyc} \frac{b^2}{b+c}.\tag{1}$$

So $$2\sum_{cyc}\frac{a^2}{b+c} \ge \sum_{cyc}\frac{a^2+b^2}{b+c}.\tag{2}$$

Now, by C-S

$$\sum_{cyc}(b+c) \sum_{cyc}\frac{a^2+b^2}{b+c} \ge \left(\sum\sqrt{a^2+b^2}\right)^2.\tag{3}$$

It's also clear that

$$\sum_{cyc}(b+c) \le \sqrt2 \sum_{cyc}\sqrt{a^2+b^2}\tag{4}.$$ From (2), (3), and (4), we conclude that $\sum \frac{a^2}{b+c}$ attains its minimum at $a=b=c$.