Find the largest real $m$, such that for all positive real $a, b, c$ with sum $1$, the inequality $\sqrt{\frac{ab} {ab+c}}+\sqrt{\frac{bc} {bc+a}}+\sqrt{\frac{ca} {ca+b}} \geq m$ is satisfied.
Problem
Source: ARO Regional stage 2023 9.9
Tags: algebra, Russia, inequalities proposed, Inequality
youthdoo
16.02.2023 15:46
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Tintarn
05.03.2023 10:25
The minimal $m$ is $1$. Indeed, the triple $(\varepsilon,\varepsilon^2,1-\varepsilon-\varepsilon^2)$ achieves values arbitrarily close to $1$ as $\varepsilon \to 0$. Now let us prove that the inequality is always true for $m=1$.
The key is to note that $ab+c=ab+(a+b+c)c=(c+a)(c+b)$. Hence we can multiply with the denominators to get the equivalent
\[\sqrt{ab(a+b)}+\sqrt{bc(b+c)}+\sqrt{ca(c+a)} \ge \sqrt{(a+b)(b+c)(c+a)}.\]Squaring and simplifying we get the equivalent
\[2a\sqrt{bc(a+b)(a+c)}+2b\sqrt{ca(b+c)(b+a)}+2c\sqrt{ab(c+a)(c+b)} \ge 2abc\]which is obvious since each of the three summands on the LHS is already at least $2abc$ (since e.g. $(a+b)(a+c)>bc$).
sqing
05.03.2023 10:56
Let $a,b,c>0$ and $a+b+c=1.$ Prove that$$\frac{3}{2}\geq \sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}+\sqrt{\frac{ab}{c+ab}}\geq 1$$https://math.stackexchange.com/questions/1632956/how-to-prove-sqrt-fracbcabc-sqrt-fraccabca-sqrt-fracabcab?noredirect=1 https://artofproblemsolving.com/community/c6h1382676p7672969 ARO (All Russian Olympiad)
Parsia--
05.03.2023 18:38
sqing wrote: Let $a,b,c>0$ and $a+b+c=1.$ Prove that$$\frac{3}{2}\geq \sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}+\sqrt{\frac{ab}{c+ab}}\geq 1$$ From the hypothesis we have: $$a+bc = a(a+b+c)+bc = (a+b)(a+c)$$$$RHS = \Sigma \frac{\sqrt{bc}}{\sqrt{(a+b)(a+c)}} \le \Sigma(\frac{b}{2a+2b} + \frac{c}{2a+2c}) = \frac{3}{2}$$
sqing
20.04.2023 05:32
Parsia-- wrote:
sqing wrote:
Let $a,b,c>0$ and $a+b+c=1.$ Prove that$$\frac{3}{2}\geq \sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}+\sqrt{\frac{ab}{c+ab}}\geq 1$$
From the hypothesis we have:
$$a+bc = a(a+b+c)+bc = (a+b)(a+c)$$$$RHS = \Sigma \frac{\sqrt{bc}}{\sqrt{(a+b)(a+c)}} \le \Sigma(\frac{b}{2a+2b} + \frac{c}{2a+2c}) = \frac{3}{2}$$
sqing wrote: Let $a,b,c>0$ and $a+b+c=1.$ Prove that$$\frac{3}{2}\geq \sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ca}{b+ca}}+\sqrt{\frac{ab}{c+ab}}\geq 1$$
Attachments:
NO_SQUARES
08.11.2023 11:03
Also, it is possible to write $\frac{a}{1-a}=x,... $ and have $z$ be some function of $x, y$ and enjoy...
NO_SQUARES
13.12.2023 23:17
I will only prove that $m=1$ works. Let $\frac{a}{1-a}=x^2, \frac{b}{1-b^2}=y^2,\frac{c}{1-c^2}=z^2$. Then we need to prove that $xy+yz+zx \geq 1$ if $\sum \frac{x^2}{x^2+1} = 1$ or $2(xyz)^2+\sum x^2y^2 = 1$ or if $2r^2+q^2-2pr = 1$ then $q \geq 1$, there $q=xy+yz+zx>0,r=xyz>0,p=x+y+z>0$. But if $q<1$ then $2r^2-2pr>0$ and so $r>p$. But obviously that $r<1$, because $r=\sqrt \frac{abc}{(a+b)(b+c)(c+a)}$. So by AM-GM $p \geq 3 \sqrt[3]{r} > \sqrt[3]{r} > r$. A conradiction!