EMC 2020 Junior Category P4 wrote: Let \(a,b,c\) be positive real numbers such that \(ab+bc+ac = a+b+c\). Prove the following inequality: \[\sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}} \leq \sqrt{2} \cdot \min \left\{ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \right\}.\] Proposed by Dorlir Ahmeti. Somewhat easy for its position, I think. Nevertheless, nice problem @dangerousliri We split the proof into two parts. Part 1: $\sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}} \leq \sqrt{2} \cdot (\frac{a}{b}+\frac{b}{c}+\frac{c}{a})$ Let $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=P$ and $a+b+c=Q$. Applying Cauchy-Schwarz we obtain $$P=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}= \frac{a^2}{ab}+\frac{b^2}{bc}+\frac{c^2}{ca} \geq \frac{(a+b+c)^2}{ab+bc+ca}=a+b+c=Q$$hence $P \geq Q$. In addition, applying Cauchy-Schwarz again, it suffices to show that $$2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})^2 \geq 3(a+b+c+\frac{a}{b}+\frac{b}{c}+\frac{c}{a})$$that is $2P^2-3P \geq 3Q$, which holds since $$3Q \leq 3P \leq 2P^2-3P$$where in the last step we used that $P \geq 3 $ by AM-GM. Part 2: $\sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}} \leq \sqrt{2} \cdot (\frac{a}{c}+\frac{b}{a}+\frac{c}{b})$ This is just a clever application of AM-GM. Indeed, $$\sqrt{2} \sum \sqrt{a+\frac{b}{c}}=\sum \sqrt{\frac{2a}{c}(c+\frac{b}{a})} \leq \sum \frac{\frac{2a}{c}+c+\frac{b}{a}}{2}=\frac{\displaystyle 3\sum \frac{a}{c}+\sum a}{2} \leq 2\sum \frac{a}{c}$$hence we are done.

square_root_of_3 wrote: Let \(a,b,c\) be positive real numbers such that \(ab+bc+ac = a+b+c\). Prove the following inequality: \[\sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}} \leq \sqrt{2} \cdot \min \left\{ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \right\}.\] Proposed by Dorlir Ahmeti. http://emc.mnm.hr/wp-content/uploads/2020/12/EMC_2020_Juniors_ENG_Solutions-2.pdf Let \(a,b,c\) be positive real numbers such that \(ab+bc+ca = a+b+c\). Prove that \[\sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}} \leq \frac{\sqrt{2}}{2} \left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c} \right).\]

Attachments:

By Cauchy inequality we have: $$ \sqrt{((ac+b) +(ba+c)+(cb+a))(\frac{1}{c} + \frac{1}{a}+\frac{1}{b})} \ge \sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}} $$So it sufficient to show that: $$ \sqrt{2} \min{(\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a}+\frac{c}{b}+\frac{a}{c})} \ge \sqrt{((ac+b) +(ba+c)+(cb+a))(\frac{1}{c} + \frac{1}{a}+\frac{1}{b})} $$This is equivalent to: $$ \sqrt{2} \min{(\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a}+\frac{c}{b}+\frac{a}{c})} \ge \sqrt{2} (a+b+c) \cdot \sqrt{\frac{1}{abc}} $$Eliminating minimum reveals that it is enough to prove: $$ a^2b+b^2c + c^2a \ge (a+b+c)\sqrt{abc} \quad \text{and} \quad ab^2+bc^2 + ca^2 \ge(a+b+c)\sqrt{abc} $$Let's prove the first one, because second one can be proved analogously. From AM -GM we have that: $$ \frac{(a^2b+ca)+(b^2c+ca) +( c^2a+ab)}{2} \ge \sqrt{a^3bc} + \sqrt{ab^3c} + \sqrt{abc^3}=(a+b+c)\sqrt{abc} $$We are left to prove that: $$ a^2b +b^2c + c^2a \ge ab+bc+ca \quad (\star) $$But by Cauchy inequality we have $(a^2b + b^2c +c^2a)(b+c+a) \ge (ab+bc+ca)^2$. Using condition $a+b+c =ab+bc+ca$, this become $(\star)$ as desired.

Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $$\sqrt{a+\frac{c}{b}}+\sqrt{b+\frac{a}{c}}+\sqrt{c+\frac{b}{a}}\leq\frac{3}{\sqrt 2}(a+b+c-1)$$$$\sqrt{a+\frac{c}{b}}+\sqrt{b+\frac{a}{c}}+\sqrt{c+\frac{b}{a}}\leq\sqrt 2(a+b+c)$$Balkan 2020 7539403284

A nice inequality! Here goes my solution which is similar to #4. EMC 2020 J4 wrote: Let \(a,b,c\) be positive real numbers such that \(ab+bc+ac = a+b+c\). Prove the following inequality: \[\sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}} \leq \sqrt{2} \cdot \min \left\{ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \right\}.\] Using the Cauchy-Schwarz inequality, and the given condition we have that$$((ac+b)+(ab+c)+(bc+a))\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right) \geq \left(\sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}}\right)^2$$$$\implies \sqrt{a+\frac{b}{c}} + \sqrt{b+\frac{c}{a}} + \sqrt{c+\frac{a}{b}}\leq \sqrt{2(a+b+c)\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right)}=\sqrt{2}(a+b+c)\sqrt{\frac{1}{abc}}$$Thus, it suffices to show that $$\min \left\{ \frac{a}{b}+\frac{b}{c}+\frac{c}{a},\ \frac{b}{a}+\frac{c}{b}+\frac{a}{c} \right\} \geq (a+b+c)\sqrt{\frac{1}{abc}}$$$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq(a+b+c)\sqrt{\frac{1}{abc}}\quad\text{and}\quad\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\geq(a+b+c)\sqrt{\frac{1}{abc}}$$It's clear that if we prove one of them the other one also follows, by multiplying the first one by $abc$ we have $$a^2c+ab^2+bc^2\geq \sqrt{a^3bc}+\sqrt{ab^3c}+\sqrt{abc^3}$$By AM-GM we have that $$\frac{(a^2c+ab)+(ab^2+bc)+(bc^2+ca)}{2}\geq \sqrt{a^3bc}+\sqrt{ab^3c}+\sqrt{abc^3}$$Thus it's enough to show that $$a^2c+ab^2+bc^2\geq\frac{(a^2c+ab)+(ab^2+bc)+(bc^2+ca)}{2}\iff a^2c+ab^2+bc^2\geq ab+bc+ca$$$$(a^2c+ab^2+bc^2)(c+a+b)\geq (ab+bc+ca)^2\implies a^2c+ab^2+bc^2\geq ab+bc+ca.\blacksquare$$