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The question is to find the minimum $K$ such that $$\sum_{cyc} \frac{a}{\sqrt{a+b}} \leq K\sqrt{a+b+c}$$ holds true for all non-negative $a,b$ and $c$.


My attempt:

I used Cauchy-Schwarz to get $$\left(\sum_{cyc} \frac{a}{\sqrt{a+b}}\right)^2=\left(\sum_{cyc} \sqrt{a}\sqrt{\frac{a}{{a+b}}}\right)^2\leq(a+b+c)\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$$

I now have to find the maximum value of $\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)$. Putting $\frac ba=x, \frac cb=y$ and $\frac ac=z$, this reduces to the problem of finding the maximum value of $$\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$$ under the constraint $xyz=1$.

I didn't get any further than this, and I'm pretty sure that my approach is wrong. For the last expression, we find that the case where $x=y=z=1$ is not the maximum as the expression is greater than $\frac32$ for $x=\frac12,y=\frac12,z=4$ (It is equal to $\frac{23}{15}$).

I found this similar question which doesn't have an answer yet.

Any help would be appreciated!

(I found this question here)

In this similar question, the maximum doesn't appear to be attained in the case where $x=y=z$.

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For $a=3$, $b=1$ and $c=0$ we obtain $k\geq\frac{5}{4}.$

We'll prove that $\frac{5}{4}$ is a minimal value.

Indeed, we need to prove that $$\sum_{cyc}\frac{a}{\sqrt{a+b}}\leq\frac{5}{4}\sqrt{a+b+c}.$$ Now, by C-S $$\sum_{cyc}\frac{a}{\sqrt{a+b}}=\sqrt{\left(\sum_{cyc}\frac{a}{\sqrt{a+b}}\right)^2}\leq\sqrt{\sum_{cyc}\frac{a}{2a+4b+c}\sum_{cyc}\frac{a(2a+4b+c)}{a+b}}.$$ Id est, it's enough to prove that $$\sum_{cyc}\frac{a}{2a+4b+c}\sum_{cyc}\frac{a(2a+4b+c)}{a+b}\leq\frac{25(a+b+c)}{16}$$ or $$2\sum_{cyc}ab(4a+b)(a+2b)^2(a-3b)^2+$$ $$+abc\sum_{cyc}(122a^4+217a^3b+143a^3c+564a^2b^2+1338a^2bc)\geq0,$$ which is obvious.

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Some thoughts

Let $x=b/a, y=c/b, z=a/c$. This transforms the inequality to

$$2≥f(x,y,z)≥1,f(x,y,z)=\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$$ Under the constraint $xyz=1$ (as you have shown).

Now \begin{align} f(x,y,z)&=\frac{(1+y)(1+z)+(1+z)(1+x)+(1+x)}{(1+y)(1+x)(1+y)(1+z)}\\ &=\frac{3+2(x+y+z)+(xy+yz+zx)}{1+(x+y+z)+(xy+yz+zx)+xyz}\\ &=\frac{3+2(x+y+z)+(xy+yz+zx)}{2+(x+y+z)+(xy+yz+zx)}\\ &=1+\left(\frac{1+(x+y+z)}{2+(x+y+z)+(xy+yz+zx)}\right)\\ &=2−\left(\frac{1+(xy+yz+zx)}{2+(x+y+z)+(xy+yz+zx)}\right). \end{align}

The term in brackets on the last two lines are non-negative. Thus, $$2≥f(x,y,z)≥1$$

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  • $\begingroup$ Hmm... yes, we can say that $2>f(x,y,z)>1$, but that contradicts the earlier equality condition in the Cauchy-Schwarz ($a=b=c$) so it gives a loose bound. $\endgroup$ – Amit Rajaraman Jul 2 at 14:33

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