$\frac{\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}}{\sqrt{2}}\leq \frac{a+b+c-1}{3}$

My solution: Observe that $\sqrt{(a+\frac{b}{c}).2} \leq \frac{a+\frac{b}{c}+2}{2}$, so we want $5(a+b+c) \geq 12+\frac{b}{c}+\frac{c}{a}+\frac{a}{b}$. Equivalently $5(a+b+c)(ab+bc+ca)\geq 36abc+3(ab^2+bc^2+ca^2)$. Equivalently $5(a^2b+b^2c+c^2a)+2(ab^2+bc^2+ca^2) \geq 21abc$, which follows from summing $2a^2b+a^2c+ac^2+2b^2c+bc^2 \geq 7abc$ because of AM-GM for 7 numbers. Edit: The above post hasn't written rightly the problem, I corrected my post before him (this was my first formulation due to a typo, and there was a latex problem so that's why he posted it, but now everything is correct)

Can you post the whole shortlist please?

A different solution. Using the fact $x\mapsto \sqrt{x}$ is concave, Jensen's inequality yields that \[ \frac13\left(\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\right)\le \sqrt{\frac13\left(a+b+c+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}. \]With this, it boils down proving (after some algebra) \[ 3(a+b+c)^2 - 8(a+b+c)+3 \ge 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right). \]We now use a fairly weak inequality: \[ ab^2+bc^2+ca^2 \le (a+b+c)(ab+bc+ca)-6abc, \]which is obvious upon expanding and applying AM-GM. Hence, it suffices to show \[ 3(a+b+c)^2-8(a+b+c)+3 \ge \frac{2(a+b+c)(ab+bc+ca)}{abc}-12 = 6(a+b+c)-12, \]where we used the fact $ab+bc+ca=3abc$. Now, set $t\triangleq a+b+c$. AM-HM inequality then yields $t\ge 3$. Under the condition $t\ge 3$, it suffices to show \[ 3t^2-8t+3 \ge 6t-12\iff 3t^2-14t+15\ge 0 \iff (3t-5)(t-3)\ge 0, \]which is clear.

It is also Azerbaijan JBMO TST Day1 P4.

By C-S, $\left(\sum \sqrt{a + \frac{b}{c}}\right)^2 \leq (\frac{1}{c} + \frac{1}{a} + \frac{1}{b})((ac+b) + (ab+c) + (bc+a)) = 3(a+b+c+ab+bc+ac)$. Squaring both sides, it suffices to show $\frac{(a+b+c-1)^2}{2} \geq \frac{1}{3}(a+b+c+ab+bc+ac)$ $\iff \frac{1}{2} (\sum a^2 + 2\sum ab - 2\sum a + 1) \geq \frac{1}{3}(\sum a + \sum ab)$ $\iff 3\sum a^2 + 6\sum ab - 6\sum a + 3 \geq 2\sum a + 2\sum ab$ $\iff 3\sum a^2 + 4\sum ab + 3 \geq 8\sum a$ $\iff 3\sum a^2 + 4\sum ab + \sum \frac{1}{a} \geq 8\sum a$ By AM-GM, $b^2 + ab + \frac{1}{a} \geq 3b$. Writing similar terms and cancelling it suffices to show $\iff 2\sum a^2 + 3\sum ab \geq 5\sum a$ Homogenizing, it suffices to show $2(\sum a^2)\frac{\sum ab}{abc} + 3\frac{(\sum ab)^2}{abc} \geq 15\sum a$ $\iff 2(\sum a^2)(\sum ab) + 3(\sum ab)^2 \geq 15\sum a^2bc$ $\iff 3\sum a^2b^2 + 2(\sum a^2)(\sum ab) \geq 9\sum a^2bc$ $\iff 3\sum a^2b^2 + 2\sum_{\text{sym}} a^3b \geq 7\sum a^2bc$ By AM-GM, $a^2b^2 + a^3b + c^3a \geq 3a^2bc , \hspace{0.25cm} b^2c^2 + b^3c + a^3b \geq 3b^2ac , \hspace{0.25cm} a^2c^2 + c^3a + b^3c \geq 3c^2ab$. Thus it suffices to show $2\sum a^2b^2+ 2\sum a^3c \geq 4\sum a^2bc$ By AM-GM, $b^2a^2 + c^3b + b^3a \geq 3b^2ac , \hspace{0.25cm} c^2b^2 + a^3c + c^3b \geq 3c^2ab , \hspace{0.25cm} a^2c^2 + b^3a + a^3c \geq 3a^2bc$ Thus it suffices to show $\sum a^2b^2 \geq \sum a^2bc$, but that is obvious by AM-GM.

Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \frac{3\sqrt 2}{3-k}(a+b+c-k)$$Where $0\leq k<3$

Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \frac{3\sqrt 2}{4}(a+b+c+1)$$$$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \frac{3\sqrt 2}{5}(a+b+c+2)$$$$\sqrt{2a+\frac{b}{c}}+\sqrt{2b+\frac{c}{a}}+\sqrt{2c+\frac{a}{b}}\leq \frac{3\sqrt 3}{3-k}(a+b+c-k)$$Where $0\leq k<3$ Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. If $$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \frac{3\sqrt 2}{k+3}(a+b+c+k),$$find the maximum of the positive real $k.$

sqing wrote: Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \frac{3\sqrt 2}{3-k}(a+b+c-k)$$Where $0\leq k<3$

sqing wrote: Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \frac{3\sqrt 2}{4}(a+b+c+1)$$$$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \frac{3\sqrt 2}{5}(a+b+c+2)$$

sqing wrote: Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \frac{3\sqrt 2}{3-k}(a+b+c-k)$$Where $0\leq k<3$ Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $$\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}\leq \sqrt{2}(a+b+c)$$ VicKmath7 wrote: Given are positive reals $a, b, c$, such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3$. Prove that $\frac{\sqrt{a+\frac{b}{c}}+\sqrt{b+\frac{c}{a}}+\sqrt{c+\frac{a}{b}}}{3}\leq \frac{a+b+c-1}{\sqrt{2}}$. $Albania$

Notice that the inequality is equivalent to: $a+b+c -1\ge \frac{\sum_{cyc}\sqrt{2(a+\frac{b}{c})}}{3}$ Furthermore by AM-GM we have that RHS $\le\frac{6+a+b+c+\sum_{cyc}\frac{a}{b}}{6}$ Which implies that we need to prove: $5\sum_{cyc}a\ge\sum_{cyc}\frac{a}{b}+12$ $\Longrightarrow 5(\sum_{cyc}a)(\sum_{cyc}ab)\ge 36abc+3\sum_{cyc}ab^2$ $\Longrightarrow 2\sum_{cyc}ab^2+5\sum_{cyc}a^2b\ge21abc$ Which clearly true by AM-GM And we are done!

First, note that by AM-HM, $a+b+c \ge 3$ (and hence, note that this implies squaring the desired inequality is reversible). Then, using the equivalent original condition $ab+bc+ca=3abc$, by AM-GM we have $abc\ge 1$, and from this it quickly follows that $a^2+b^2+c^2\ge ab+bc+ca\ge 3$. (These will be useful later on.) By Holder, we have that \[\left(\sum_{cyc} \sqrt{\frac{ac+b}{c}}\right)^2 \le \left(\sum_{cyc}ac+b\right)\left(\sum_{cyc} \frac{1}{c}\right) = 3\sum_{cyc}(ac+b).\]Thus, it suffices to prove that \[\sum_{cyc}(ac+b) \le \frac{3}{2}(a+b+c-1)^2.\]Let $s=a+b+c$. Expanding, rearranging, and multiplying the inequality through by $2$, we wish to prove that \[0\le 2s^2-8s+3+(a^2+b^2+c^2).\]The minimum of $2s^2-8s+3$ occurs at $s=2$; however, recall that $s\ge 3$. Thus, $2s^2-8s+3\ge 2(3)^2-8(3)+3 = -3$. Since $a^2+b^2+c^2\ge 3$, we are done. $\blacksquare$

Hmm another inequality By the tangent line trick we have that $\sqrt{x}\le\frac{1}{2\sqrt{2}}(x-2)+\sqrt{2}=\frac{1}{2\sqrt{2}}x+\frac{1}{\sqrt{2}}$ because of $\sqrt{x}$’s concavity, so we have that: \begin{align*} \frac{1}{3}\sum_\textrm{cyc}\sqrt{a+\frac{b}{c}}&\le\frac{1}{3}\sum_\textrm{cyc}\left(\frac{1}{2\sqrt{2}}\left(a+\frac{b}{c}\right)+\frac{1}{\sqrt{2}}\right)\\ &=\frac{1}{\sqrt{2}}+\frac{1}{6\sqrt{2}}\left(a+b+c+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right) \end{align*}so it suffices to show that \begin{align*} \frac{1}{\sqrt{2}}+\frac{1}{6\sqrt{2}}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+\frac{1}{6\sqrt{2}}(a+b+c)&\le\frac{1}{\sqrt{2}}(a+b+c-1)\\ 2+\frac{1}{6}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)&\le\frac{5}{6}(a+b+c)\\ 12+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}&\le5(a+b+c) \end{align*}So now that we’ve moved to this simplified version we are going to dehomogenise the whole inequality; seeing as there are two degree $0$ terms in the LHS and only one degree $1$ term in the RHS with the condition being degree $-1$, we can then see it suffices to show that: \begin{align*} 12+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}&\le\frac{5}{3}(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\\ 21&\le2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+5\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right) \end{align*}However the above inequality is almost trivially true; indeed, take AM-GM on the terms within the brackets, and obtain $2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\ge2\cdot\sqrt[3]{\left(\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}\right)}=6$, and $5\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\ge5\cdot3\sqrt[3]{\left(\frac{b}{a}\cdot\frac{c}{b}\cdot\frac{a}{c}\right)}=15$, as desired.

slightly different to the above ones