[moved from http://www.artofproblemsolving.com/Forum/posting.php?mode=edit&f=52&p=1933051] Hi, I'm very glad to see that my problem was included in IMO Shortlist , because I was in Spain yet as a contestant. Originally I proposed this problem with the following proof. As I now this version was sent by out team leader without any corrections to IMO 2009 Problem Selection Committee: We have \[\sqrt{x^2+y^2} \leq \sqrt{2}(x+y-\sqrt{xy}) \quad \forall x ,y >0 \eqno (*),\] because $(*) \Leftrightarrow x^2+y^2 \leq 2(x^2+y^2+xy +2xy-2x\sqrt{xy}-2y\sqrt{xy}) \Leftrightarrow (\sqrt{x}-\sqrt{y})^4 \geq 0,$ which is clearly. Therefore \[\sqrt{\frac{x^2+y^2}{x+y}} \leq \frac{\sqrt{2}(x+y-\sqrt{xy})}{\sqrt{x+y}}=\sqrt{2(x+y)}-\frac{\sqrt{2xy}}{\sqrt{x+y}} \quad \forall x ,y >0. \] This implies that \[\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}} \leq \sqrt{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})-\sqrt{2}\left( \frac{\sqrt{ab}}{\sqrt{a+b}}+ \frac{\sqrt{bc}}{\sqrt{b+c}}+\frac{\sqrt{ca}}{\sqrt{c+a}} \right). \] Then, it is enough to prove that \[\frac{\sqrt{ab}}{\sqrt{a+b}}+ \frac{\sqrt{bc}}{\sqrt{b+c}}+\frac{\sqrt{ca}}{\sqrt{c+a}} \geq \frac{3}{\sqrt{2}}.\] Denote by $x=\frac{1}{a},$ $y=\frac{1}{b},$ и $z=\frac{1}{c}.$ We have $ab+bc+ca \leq 3abc \Leftrightarrow \frac{1}{a}+ \frac{1}{b}+\frac{1}{c} \leq 3 \Leftrightarrow x+y+z \leq 3.$ It is easy to see that \[ \frac{\sqrt{ab}}{\sqrt{a+b}}+ \frac{\sqrt{bc}}{\sqrt{b+c}}+\frac{\sqrt{ca}}{\sqrt{c+a}}=\frac{1}{\sqrt{\frac{1}{a}+\frac{1}{b}}}+ \frac{1}{\sqrt{\frac{1}{b}+\frac{1}{a}}}+\frac{1}{\sqrt{\frac{1}{c}+\frac{1}{a}}}=\frac{1}{\sqrt{x+y}}+ \frac{1}{\sqrt{y+z}}+\frac{1}{\sqrt{z+x}} \geq \\ \left[ AM-HM \right] \geq \frac{9}{\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}} \geq \left[ AM-QM \right] \geq \frac{9}{\sqrt{3((x+y)+(y+z)+(z+x))}}=\\ \frac{9}{\sqrt{6(x+y+z)}}=\left[ x+y+z \leq 3 \right] \geq \frac{9}{\sqrt{6 \cdot 3}}=\frac{3}{\sqrt{2}}, \] and we are done. Equality occurs if and only if $a=b=c=1.$ P.s. The best way to prove (*) is to say that it's equiv to the inequality $\sqrt{x^{2}+{y^2}}+\sqrt{2xy} \leq \sqrt{2(x+y)^2}=\sqrt{2}(x+y)$, which is true by AM-QM.

Let $x = \frac{1}{a}$, $y = \frac{1}{b}$, and $z = \frac{1}{c}$. We wish to show that if $x + y + z \leq 3$, then $3 + \sum_{\mbox{cyc}} \sqrt{\frac{1}{x} + \frac{1}{y} - \frac{2}{x+y}} \leq \sqrt{2} \left(\sum_{\mbox{cyc}} \sqrt{\frac{1}{x} + \frac{1}{y}} \right).$ By the power mean inequality on $\sqrt{\frac{2}{x+y}} + \sqrt{\frac{2}{y+z}} + \sqrt{\frac{2}{z+x}}$, the 1st power mean of these numbers is greater than or equal to the -2nd power mean, whence $\sqrt{\frac{2}{x+y}} + \sqrt{\frac{2}{y+z}} + \sqrt{\frac{2}{z+x}} \geq \frac{3 \sqrt{3}}{\sqrt{x+y+z}}$. Rearrange this to get $\frac{3 \sqrt{3}}{\sqrt{x+y+z}} + \sum_{\mbox{cyc}} \left( \sqrt{2\left(\frac{1}{x} + \frac{1}{y} \right)} - \sqrt{\frac{2}{x+y}} \right) \leq \sqrt{2} \left(\sum_{\mbox{cyc}} \sqrt{\frac{1}{x} + \frac{1}{y}} \right).$ Note that whenever $m \geq n$, $\sqrt{2m} - \sqrt{n} \geq \sqrt{m-n} \iff (\sqrt{m} - \sqrt{2n})^2$. Furthermore, observe that $\frac{3 \sqrt{3}}{\sqrt{x+y+z}} \geq 3$. Applying these two inequalities gives \begin{align*} 3 + \sum_{\mbox{cyc}} \sqrt{\frac{1}{x} + \frac{1}{y} - \frac{2}{x+y}} &\leq \frac{3 \sqrt{3}}{\sqrt{x+y+z}} + \sum_{\mbox{cyc}} \left( \sqrt{2\left(\frac{1}{x} + \frac{1}{y} \right)} - \sqrt{\frac{2}{x+y}} \right) \\ &\leq \sqrt{2} \left(\sum_{\mbox{cyc}} \sqrt{\frac{1}{x} + \frac{1}{y}} \right), \end{align*} which was what we wanted.

This also appeared in Indian IMO Training Camp 2010; see here and I also proposed it at the Inequalities Marathon, where I gave another solution to this extremely nice problem.

(1) By Power-Mean inequality, \[\sqrt{2} \cdot \sqrt{a+b}=\sqrt{2} \cdot \sqrt{\frac{(a+b)^2}{a+b}}=\sqrt{2} \cdot \sqrt{\frac{a^2 + b^2}{a+b}+\frac{2ab}{a+b}} \ge \sqrt{\frac{a^2 + b^2}{a+b}}+\sqrt{\frac{2}{\frac{1}{a}+\frac{1}{b}}} \] (2) By AM-HM, \[\sqrt{\frac{2}{\frac{1}{a}+\frac{1}{b}}} +\sqrt{\frac{2}{\frac{1}{b}+\frac{1}{c}}} +\sqrt{\frac{2}{\frac{1}{c}+\frac{1}{a}}} \ge \frac{9\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}}+\sqrt{\frac{1}{b}+\frac{1}{c}}+\sqrt{\frac{1}{c}+\frac{1}{a}}} \] By Power-Mean inequality, \[\frac{9\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}}+\sqrt{\frac{1}{b}+\frac{1}{c}}+\sqrt{\frac{1}{c}+\frac{1}{a}}} \ge \frac{9\sqrt{2}}{\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}}} \ge 3\] By the condition that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \le 3$. Summing inequality (1) for all pairs in $\{ a,b,c \}$ and by inequality (2), \[ \sqrt{\frac{a^{2}+b^{2}}{a+b}}+\sqrt{\frac{b^{2}+c^{2}}{b+c}}+\sqrt{\frac{c^{2}+a^{2}}{c+a}}+3\leq\sqrt{2}\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right) \quad \blacksquare \]

By AM-GM, $\sqrt{x}+\sqrt{y} \le \sqrt{2(x+y)}$ $$\implies \sum \sqrt{2(a+b)} \ge \sum \sqrt{\frac{a^2+b^2}{a+b}} + \sum \sqrt{\frac{2ab}{a+b}}$$ Thus we are left to show $\sum \sqrt{\frac{2ab}{a+b}} \ge 3$ Let $S =\sum \sqrt{\frac{2ab}{a+b}}$ By Holder's $$S^2(\sum \frac{1}{2a}+\frac{1}{2b})\ge 27$$ $$S^2(\sum \frac{1}{a}) \ge 27$$ Thus by the given condition, $$S^2 \ge 9$$ $$S \ge 3$$ Hence proved.

at first $ 1+\sqrt{\frac{x^2+y^2}{x+y}} \leq \sqrt{2} \times \sqrt{1+\frac{x^2+y^2}{x+y}} $ so,we only need to prove that $ \Sigma \sqrt{1+\frac{a^2+b^2}{a+b}} \leq \Sigma \sqrt{a+b} $ and we have $ 1+\frac{a^2+b^2}{a+b}=a+b+1-\frac{2}{\frac{1}{a}+\frac{1}{b}} $ so $ 1+\frac{a^2+b^2}{a+b} \geq a+b $ if and only if $ \frac{1}{a}+\frac{1}{b} \geq 2 $ WLOG , $ a \geq b \geq c $ , then $ 1+\frac{a^2+b^2}{a+b} \leq a+b $ $ 1+\frac{b^2+c^2}{b+c} \geq b+c $ now prove $ \Sigma (1+\frac{a^2+b^2}{a+b}) \leq \Sigma (a+b) $ <=> $ 3 \leq \Sigma \frac{2}{\frac{1}{a}+\frac{1}{b}} $ it is right because of $ \frac{3}{\Sigma \frac{2}{\frac{1}{a}+\frac{1}{b}}} \leq \frac{\Sigma \frac{1}{a}}{3} \leq 1 $ by karamata , done!

Quite an awesome inequality! IMO SL 2009 A4 wrote: Let $a$, $b$, $c$ be positive real numbers such that $ab+bc+ca\leq 3abc$. Prove that \[\sqrt{\frac{a^2+b^2}{a+b}}+\sqrt{\frac{b^2+c^2}{b+c}}+\sqrt{\frac{c^2+a^2}{c+a}}+3\leq \sqrt{2}\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)\] Proposed by Dzianis Pirshtuk, Belarus Notice that for all real numbers $x>y>0$ we have $$\sqrt{x-y} \le \sqrt{2x}-\sqrt{y} \iff x^2 \ge 4y(x-y) \iff (x-2y)^2 \ge 0.$$Write $$\sqrt{\frac{x^2+y^2}{x+y}}=\sqrt{(x+y)-\frac{2xy}{x+y}} \le \sqrt{2(x+y)}-\sqrt{\frac{2xy}{x+y}},$$so our given inequality is equivalent to $$\sum_{\text{cyc}} \sqrt{\frac{2ab}{a+b}} \ge 3.$$Apply Holder's inequality to get $$\left(\sum_{\text{cyc}} \sqrt{\frac{2ab}{a+b}}\right)^2 \left(\sum_{\text{cyc}} \frac{a+b}{2ab} \right) \ge 27.$$However, since $$\sum_{\text{cyc}} \frac{a+b}{2ab}=\sum_{\text{cyc}} \frac{1}{a} \le 3,$$we get the desired bound. Equality occurs when $a=b=c=1$. $\blacksquare$

Let $x = \frac 1a$ etc., then $x+y+z \leq 3$. We wish to show \[ \sum \sqrt{2(a+b)} - \sqrt{\frac{a^2+b^2}{a+b}} \geq 3 \iff \sum \frac{\sqrt 2(a+b)-\sqrt{a^2+b^2}}{\sqrt{a+b}} \geq 3 \iff \sum \frac{\sqrt{2} (x+y) - \sqrt{x^2+y^2}}{\sqrt{xy(x+y)}} \geq 3. \]Note that multiplying $x,y,z$ by $\lambda \in [1, \infty)$ does not increase the left hand side, so assume $x+y+z = 3$. Now, we claim that in fact \[ \sqrt 2(x+y)-\sqrt{x^2+y^2} \geq \sqrt{2xy}. \]Indeed, let $x = u^2, y = v^2$ for $u,v \geq 0$, then we want \[ \sqrt{2}(u^2+v^2-uv) \geq \sqrt{u^4+v^4}, \]or equivalently \[ \sqrt 2((u-v)^2+uv) \geq \sqrt{u^4+v^4} \iff 2((u-v)^2+uv)^2 \geq u^4+v^4 \iff 2(u-v)^2((u-v)^2+2uv) \geq (u^2-v^2)^2 = (u-v)^2(u+v)^2. \]If $u=v$ we are clearly done, else we want to show $2u^2+2v^2 \geq (u+v)^2$ which is clear. So, it suffices to show \[ \sum \frac{\sqrt 2}{\sqrt{x+y}} \geq 3\iff \sum \sqrt{\frac{2}{3-x}} \geq 3. \]However, $(3-x)^{-\frac 12}$ is convex with second derivative $\frac{3}{4}(3-x)^{-\frac 52}$, so we have \[ \sum \frac{\sqrt 2}{\sqrt{3-x}} \geq \frac{\sqrt 2\cdot 3}{\sqrt{3-1}} = 3 \]as desired. $\blacksquare$

Note, $a,b,c>0$ $\sqrt {a+b}={\sqrt {\frac {(a+b)^2}{(a+b)}}}={\sqrt {\frac {a^2+b^2}{a+b}+\frac {2ab}{a+b}}}\implies\geqslant \frac {1}{\sqrt 2}\left (\sqrt {\frac{a^2+b^2}{a+b}}+\sqrt {\frac {2ab}{a+b}}\right)\implies\sqrt {2(a+b)}$ $\ge \sum{\sqrt {\frac {a^2+b^2}{a+b}}}+\sum {\sqrt {\frac {2ab}{a+b}}}=\sum {\sqrt {\frac {(a^2+b^2)}{a+b}}}+\sum\frac {1}{\sqrt {\frac {1}{2a}+\frac {1}{2b}}}\ge \sum {\sqrt {\frac {(a^2+b^2)}{a+b}}}+\frac {9}{\sqrt {3\left (\frac {1}{a}+\frac {1}{b}+\frac {1}{c}\right)}}$ $=\sum {\sqrt {\frac {(a^2+b^2)}{a+b}}}+\frac {9}{\sqrt {3\frac {\sum ab}{abc}}}\ge{\sqrt {\frac {(a^2+b^2)}{a+b}}}+3$

A very nice problem indeed! We first prove a well-known inequality. Claim: $\sqrt{x}+\sqrt{y}\leq \sqrt{2(x+y)}$. Proof: The claim follows directly from AM-GM. We have that $x+y\geq 2\sqrt{xy}\leftrightarrow 2x+2y\geq x+y+2\sqrt{xy}=(\sqrt{x}+\sqrt{y})^2\leftrightarrow \sqrt{x}+\sqrt{y}\leq \sqrt{2(x+y)}.$ From the claim, we have that $$\sum_{cyc} \sqrt{2(a+b)}\geq \sum_{cyc} \sqrt{\frac{a^2+b^2}{a+b}}+\sum_{cyc}\sqrt{\frac{2ab}{a+b}}.$$ Hence, in order to prove the problem, we just have to show that $\sum_{cyc} \sqrt{\frac{2ab}{a+b}}\geq 3$. Luckily, this is easy by Holder: $$(\sum_{cyc} \sqrt{\frac{2ab}{a+b}})^2(\sum_{cyc} \frac{4a+4b}{ab})\geq (2+2+2)^3=6^3\implies$$$$\sum_{cyc} \sqrt{\frac{2ab}{a+b}}\geq \sqrt{\frac{216}{4(2)(3)}}=3.$$ We are done!

By Hölder: $$(\sum_{\text{cyc}} \sqrt{\frac{2ab}{a+b}}) (\sum_{\text{cyc}} \sqrt{\frac{2ab}{a+b}}) (\sum_{\text{cyc}} \frac{a+b}{2ab})\geq 27$$$$\iff (\sum_{\text{cyc}} \sqrt{\frac{2ab}{a+b}})(\sum_{\text{cyc}} \sqrt{\frac{2ab}{a+b}}) (\sum_{\text{cyc}}\frac{1}{a}) \geq 27$$$$\iff (\sum_{\text{cyc}} \sqrt{\frac{2ab}{a+b}}) \geq 3$$By AM-QM: $$\sum_{\text{cyc}} \sqrt{2}\sqrt{a+b} \geq \sum_{\text{cyc}} \sqrt{\frac{2a}{a+b}} +\sum_{\text{cyc}} \sqrt{\frac{a^2+b^2}{a+b}}$$Q.E.D.

Let $x=\frac1a$, $y=\frac1b$, and $z=\frac1c$. Then, $x+y+z\leq3$, so the inequality is equivalent to $$\sqrt{\frac{x^2+y^2}{xy(x+y)}}+\sqrt{\frac{y^2+z^2}{yz(y+z)}}+\sqrt{\frac{z^2+x^2}{zx(z+x)}}+3\leq\sqrt2\left(\sqrt{\frac{x+y}{xy}}+\sqrt{\frac{y+z}{yz}}+\sqrt{\frac{z+x}{zx}}\right),$$or $$\sqrt{\frac{2x+2y}{xy}}-\sqrt{\frac{x^2+y^2}{xy(x+y)}}+\sqrt{\frac{2y+2z}{yz}}-\sqrt{\frac{y^2+z^2}{yz(y+z)}}+\sqrt{\frac{2z+2x}{zx}}-\sqrt{\frac{z^2+x^2}{zx(z+x)}}\geq3.$$This rearranges to $$\frac1{\sqrt{xy(x+y)}}\left(\sqrt2(x+y)-\sqrt{x^2+y^2}\right)+\frac1{\sqrt{yz(y+z)}}\left(\sqrt2(y+z)-\sqrt{y^2+z^2}\right)+\frac1{\sqrt{zx(z+x)}}\left(\sqrt2(z+x)-\sqrt{z^2+x^2}\right)\geq3.$$By QM-AM, we get $\sqrt{2xy}+\sqrt{x^2+y^2}\leq\sqrt2(x+y)$, which implies $\sqrt2(x+y)-\sqrt{x^2+y^2}\geq\sqrt{2xy}$. Therefore, we get that the left hand side is equal to $$\sqrt{\frac2{x+y}}+\sqrt{\frac2{y+z}}+\sqrt{\frac2{z+x}}.$$This means that by Holder's Inequality, we have \begin{align*} \sqrt{\frac2{x+y}}+\sqrt{\frac2{y+z}}+\sqrt{\frac2{z+x}}&=\frac{\left(2^{\frac13}\right)^{\frac32}}{(x+y)^{\frac12}}+\frac{\left(2^{\frac13}\right)^{\frac32}}{(y+z)^{\frac12}}+\frac{\left(2^{\frac13}\right)^{\frac32}}{(z+x)^{\frac12}}\\ &\geq\frac{\left(2^{\frac13}+2^{\frac13}+2^{\frac13}\right)^{\frac32}}{(2x+2y+2z)^{\frac12}}\\ &=\frac{3^{\frac32}2^{\frac12}}{2^{\frac12}(x+y+z)^{\frac12}}\\ &=\frac{3^{\frac32}}{(x+y+z)^{\frac12}}\\ &\geq3. \end{align*}

Note that $a+b+2\sqrt{ab}\le 2a+2b$ which means $\sqrt{a}+\sqrt{b}\ge\sqrt{2}\le \sqrt{a+b}$, so \[\sqrt{\frac{a^2+b^2}{a+b}}\le \sqrt{2}\left(\sqrt{a+b}-\sqrt{\frac{2ab}{a+b}}\right)\]so it remains to show that $\left(\sum_{cyc}\sqrt{\frac{2ab}{a+b}}\right)\le 3$ By Holder's \[\left(\sum_{cyc}\sqrt{\frac{2ab}{a+b}}\right)^2(\sum_{cyc}\frac{a+b}{2ab})\ge 27\], but $\sum_{cyc}\frac{a+b}{2ab}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le 3$ which implies the result.

There was generalization 1 here.

There was generalization 2 here.

Wow, I actually have a solution which is pretty clean!? Fudging + Jensen’s Inequality Homogenizing, it suffices to show that \[3\sqrt{3} \cdot \sqrt{\frac{abc}{ab + bc + ca}} + \sum_{\mathrm{cyc}} \sqrt{\frac{a^2 + b^2}{a + b}} \le \sqrt{2}(\sqrt{a + b} + \sqrt{b + c} + \sqrt{c + a}).\] By Jensen's Inequality on $f(x) = x^{-1/2}$ which is convex for positive reals $x$, we get that \[3\sqrt{\frac{3}{ab + bc + ca}} \le \sum_{\mathrm{cyc}} \sqrt{\frac{2}{ac + bc}}.\]By multiplying both sides by $\sqrt{abc}$, this becomes \[3\sqrt{3} \cdot \sqrt{\frac{abc}{ab + bc + ca}} \le \sum_{\mathrm{cyc}} \sqrt{\frac{2ab}{a + b}}.\]Thus, it suffices to show that \[\sqrt{\frac{2ab}{a + b}} + \sqrt{\frac{a^2 + b^2}{a + b}} \le \sqrt{2(a + b)}.\]This follows by Jensen's Inequality on $f(x) = x^{1/2}$, which is concave for positive reals $x$. Thus, we are done.

This is going to be similar to above solutions At least I solved an inequality Note by power mean with $r=\frac{1}{2},1$: \begin{align*} \sqrt{\frac{2ab}{a+b}}+\sqrt{\frac{a^2+b^2}{a+b}}&\le2\sqrt{\frac{1}{2}\left(\frac{2ab}{a+b}+\frac{a^2+b^2}{a+b}\right)}\\ &=\sqrt{2}\sqrt{\frac{(a+b)^2}{a+b}}\\ &=\sqrt{2}\sqrt{a+b} \end{align*}So summing cyclically: \begin{align*} \sum_{\textrm{cyc}}\sqrt{\frac{2ab}{a+b}}+\sqrt{\frac{a^2+b^2}{a+b}}&\le\sqrt{2}\sum_{\textrm{cyc}}\sqrt{a+b} \end{align*}So it suffices to end up showing that $\sum_{\textrm{cyc}}\sqrt{\frac{2ab}{a+b}}\ge3$. This is where we start using the condition. (!) Since I don’t like dealing with $ab+bc+ca$ and $abc$ all in one go, we are going to substitute $x=\frac{1}{a}$, etc, so the condition reduces down to $x+y+z\le3$ whilst the inequality desired reduces down to \begin{align*} \sum_{\textrm{cyc}}\sqrt{\frac{2}{x+y}}&\ge3 \end{align*}Now notice $x+y\le 3-z$ so it suffices to show that \begin{align*} \sum_{\textrm{cyc}}\sqrt{\frac{2}{3-x}}&\ge3 \end{align*}However this is just Jensen with $f(x)=\sqrt{\frac{2}{3-x}}$ or tangent line trick at $x=1$; the actual calculus computation is deferred to desmos (again!) Thus done