mattilgale wrote: If $ a$, $ b$, $ c$ are the sides of a triangle, prove that \[ \sum_{\mbox{cyc}}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \] $ \sum_{\mbox{cyc}}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3\Leftrightarrow\sum_{cyc}\left(1-\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)\geq0\Leftrightarrow$ $ \Leftrightarrow\sum_{cyc}\frac{\sqrt{ab}-\sqrt{c(a+b-c)}}{\sqrt a+\sqrt b-\sqrt c}\geq0\Leftrightarrow$ $ \Leftrightarrow\sum_{cyc}\frac{(a-b)(a-c)}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)}\geq0.$ Let's $ a\geq b\geq c.$ Then $ \frac{1}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)}\geq\frac{1}{(\sqrt{ac}+\sqrt{b(a+c-b)})(\sqrt a+\sqrt c-\sqrt b)}$ since, $ \sqrt{ac}+\sqrt{b(a+c-b)}\geq\sqrt{bc}+\sqrt{a(b+c-a)}$ and $ \sqrt a+\sqrt c-\sqrt b\geq\sqrt b+\sqrt c-\sqrt a.$ Thus, your inequality is proved.

My solution is the same with yours

I don't understand this one Quote: $\sum_{cyc}\left(1-\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)\geq0 \Leftrightarrow\sum_{cyc}\frac{\sqrt{ab}-\sqrt{c(a+b-c)}}{\sqrt a+\sqrt b-\sqrt c}\geq 0$

zibi wrote: I don't understand this one Quote: $\sum_{cyc}\left(1-\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)\geq0 \Leftrightarrow\sum_{cyc}\frac{\sqrt{ab}-\sqrt{c(a+b-c)}}{\sqrt a+\sqrt b-\sqrt c}\geq 0$ You are right: $\sum_{cyc}\left(1-\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\right)\geq0 \Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{\sqrt{ab}-\sqrt{c(a+b-c)}}{\left(\sqrt a+\sqrt b-\sqrt c\right)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{a+b-c}\right)}\geq 0\Leftrightarrow$ $\Leftrightarrow\sum_{cyc}\frac{(a-b)(a-c)}{(\sqrt{bc}+\sqrt{a(b+c-a)})(\sqrt b+\sqrt c-\sqrt a)\left(\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}\right)}\geq0.$ But $\sqrt a+\sqrt b+\sqrt c+\sqrt{a+c-b}\geq\sqrt a+\sqrt b+\sqrt c+\sqrt{b+c-a}$ true too. Thank you.

What was the idea behind the first (mistaken) simplification?

mattilgale wrote: If a, b, c are sides of a triangle prove that \[\sum_{\mbox{cyc}}\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \] $\sum \frac{\sqrt{-a+b+c}}{-\sqrt{a}+\sqrt{b}+\sqrt{c}}$ $\leq \sqrt{3\sum \frac{-a+b+c}{(-\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}}$ by Cauchy's $=\sqrt{3\sum \frac{2x^{2}+2xy+2xz-2yz}{4x^{2}}}$ (where $x=\frac{-\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}$, and so on) $=\sqrt{3\sum \frac{x^{2}+xy+xz-yz}{2x^{2}}}$ $=\sqrt{3 \frac{3x^{2}y^{2}z^{2}+\sum_{sym}x^{3}y^{2}z-\sum_{cyc}x^{3}y^{3}}{2x^{2}y^{2}z^{2}}}$ $\leq \sqrt{3 \frac{6x^{2}y^{2}z^{2}}{2x^{2}y^{2}z^{2}}}=3$ by Schur's for $xy,yz,zx$.

Let me post the solution I had for this problem (I was hoping that I would get to coordinate this problem if it were chosen in the competition last year ) Since $ a,b,c$ are the sides of a triangle, it is clear that $ x=\sqrt a, y=\sqrt b, z= \sqrt c$ are also the sides of a triangle (so the numbers $ -x+y+z$, $ x-y+z$ and $ x+y-z$ are positive). We first apply a Cauchy to get rid of the square roots in the numerators of the fractions \[ \left(\sum \frac{ \sqrt{-x^{2}+y^{2}+z^{2}}}{-x+y+z}\right)^{2}\leq 3 \sum \frac{-x^{2}+y^{2}+z^{2}}{(-x+y+z)^{2}}.\] Therefore what we want to prove is \[ \sum \frac{-x^{2}+y^{2}+z^{2}}{(-x+y+z)^{2}}\leq 3. \] Moving all into one side (and distributing the 3 to each of the fractions equally) we get \[ 0 \leq \sum \frac{ x^{2}+yz-xy-xz }{ (-x+y+z)^{2}}= \sum \frac 1{(-x+y+z)^{2}}(x-y)(x-z) , \] which of course is nothing else than Vornicu-Schur, because if $ x \geq y\geq z$ then $ \frac 1{(-x+y+z)^{2}}\geq \frac 1{(x-y+z)^{2}}\geq \frac 1{(x+y-z)^{2}}$ and we are done.

Who is the proposer of that problem ? It's from Korea according to IMO compedium so Hojoo Lee ?

silouan wrote: Who is the proposer of that problem ? It's from Korea according to IMO compedium so Hojoo Lee ? According to http://ultrametric.googlepages.com/proposals.pdf this problem indeed has been constructed by Hojoo Lee.

Is It True ? [tex]$ \frac{\sqrt[n]{a+b-c}}{\sqrt[n]{a}+\sqrt[n]{b}-\sqrt[n]{c}}+ \frac{\sqrt[n]{b+c-a}}{\sqrt[n]{b}+\sqrt[n]{c}-\sqrt[n]{a}}+ \frac{\sqrt[n]{c+a-b}}{\sqrt[n]{c}+\sqrt[n]{a}-\sqrt[n]{b}}\leq 3$ ?

Put $ a=x+y, b=x+z, c=y+z$. Then we should prove that $ \sum\frac{\sqrt{2x}}{\sqrt{x+y}+\sqrt{x+z}-\sqrt{y+z}}\le 3$. By Cauchy, we get that $ LHS\le \frac{(\sqrt[4]{2x}+\sqrt[4]{2y}+\sqrt[4]{2z})^2} {\sqrt{x+y}+\sqrt{x+z}+\sqrt{y+z}}\le 3\frac{\sqrt{2x}+\sqrt{2y}+\sqrt{2z}} {\sqrt{x+y}+\sqrt{x+z}+\sqrt{y+z}}= 3\frac{2\sqrt{x}+2\sqrt{y}+2\sqrt{z}} {\sqrt{2x+2y}+\sqrt{2x+2z}+\sqrt{2y+2z}}$. $ \sqrt{x}+\sqrt{y}\le \sqrt{2x+2y}$, because its squared is true: $ x+y+2\sqrt{xy}\le x+y+x+y$ (we have positive numbers here), so numerator$ \le$denominator. Thus $ 3\frac{2\sqrt{x}+2\sqrt{y}+2\sqrt{z}} {\sqrt{2x+2y}+\sqrt{2x+2z}+\sqrt{2y+2z}}\le 3$. We're done.

Uh.. sorry, my solution above is wrong..

we take $ a=x^{2}$ $ b=y^{2}$ $ c=z^{2}$ then $ x,y,z$ are also side of triangle so there are exist uniqe triple such that $ x=a'+b' y=b'+c' z=c'+a'$ so only condition for triples $ (a',b',c')$ should be $ a',b',c' \in R^+$ the we rewrite our inequality $ \sum\frac{\sqrt{4a'^2-2(a'-b')(a'-c')}}{2a'}$ but we have $ 4a'^2-2(a'-b')(a'-c')$ is negatif for example triple $ (1,10,2)$ where is my fault? can anyone tell me

I find my mistake.I use Ozandro's Lemma for this solution but it is not true

mestav wrote: I find my mistake.I use Ozandro's Lemma for this solution but it is not true what is Ozandro's Lemma?

other solution: We have that \[ \frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}} =\sqrt{1-2\frac{(\sqrt{c}-\sqrt{a})(\sqrt{c}-\sqrt{b})}{(\sqrt{a}+\sqrt{b}-\sqrt{c})^2}} \le 1-\frac{(\sqrt{c}-\sqrt{a})(\sqrt{c}-\sqrt{b})}{(\sqrt{a}+\sqrt{b}-\sqrt{c})^2} .\] So we prove that if $x=\sqrt{b}+\sqrt{c}-\sqrt{a}>0$, $y=\sqrt{a}+\sqrt{c}-\sqrt{b}>0$, $z=\sqrt{a}+\sqrt{b}-\sqrt{c}>0$ then \[ \sum x^{-2}(x-y)(x-z) \ge 0. \] It is obviously true by Schur's inequality.

Hey mathuz Schur's Inequality tells $r>0$ but you have $r=-2$

TripteshBiswas wrote: Hey mathuz Schur's Ineuality tells r>0 but you have r=-2 See here the proof when $r$ is negative: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=415872 the last post

thanks mudok it was not known to me when r is negative

Lemma: $\sum_{cyc} \frac {(x-z)(x-y)}{x^2}\ge 0$. Since $2(x-y)(x-z)=(x-y)^2+(x-z)^2-(y-z)^2$ the inequality is reduced to the inequality $\sum_{cyc} (x-y)^2(\frac {1}{x^2}+\frac {1}{y^2}-\frac{1}{z^2})$ where we deonte $S_x=\frac {1}{x^2}+\frac {1}{y^2}-\frac{1}{z^2}$ this is indeed true by the SOS method just assume WLOG $x\ge y\ge z$ and put $x-y=a,y-z=b$ the inequality becomes $a^2(S_z+S_y)+b^2(S_x+S_y)+2abS_y$ but we can easily see that $S_z+S_y,S_x+S_y,S_y$ are all greater than $0$ so our inequality is proven. Back to the problem Denote by $S_a,S_b,S_c$ those expressions in the inequality we need to prove. Now notice that the inequality $\sum_{cyc}(1-(S_a)^2)\ge 0$ reduces to the lemma for numbers $\sqrt{a}+\sqrt{b}-\sqrt{c},\sqrt{a}-\sqrt{b}+\sqrt{c},-\sqrt{a}+\sqrt{b}+\sqrt{c}$. This gives $\sum_{cyc} (S_a)^2\le 3$ and from here we easily get the desiret result

My solution is easy Since $a,b,c$ are the sides of a triangle, we can set $a=(q+r)^2$, $b=(p+r)^2$, $c=(p+q)^2$ Thus the inequality can be rewritten as $\sum_{cyclic} \frac{\sqrt{2p^2+2pr+2pq-2qr}}{p} \le 6$ Thus is it is sufficient to show that $\sum_{cyclic} \frac{2p^2+2pr +2pq -2qr}{p^2} \le 12$ $\iff \sum_{cyclic} \frac{pr+pq-qr}{p^2} \le 3$ $\iff \sum_{cyclic} q^2r^2(pr+pq-qr) \le 3p^2q^2r^2$ This is third degree Schur's inequality for $pq,qr,rp$ Hence proved

Let a,b,c>0,and a+b+c=3 prove that: ${\frac {a\sqrt {\sqrt {b}+\sqrt {c}}}{\sqrt {a+\sqrt {bc}}}}+{\frac {b \sqrt {\sqrt {c}+\sqrt {a}}}{\sqrt {b+\sqrt {ac}}}}+{\frac {c\sqrt { \sqrt {a}+\sqrt {b}}}{\sqrt {c+\sqrt {ab}}}}\leq 3.$ http://www.artofproblemsolving.com/community/c6h1098581p4979124

mattilgale wrote: If $a,b,c$ are the sides of a triangle, prove that \[\frac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\leq 3 \]Proposed by Hojoo Lee, Korea TST PERU - IMO 2007

my solution: consider a function $ f(x)=\sqrt{x} $ and three point A(a,$ \sqrt{a} $),B(b,$ \sqrt{b} $),C(c,$ \sqrt{c} $) choose D,E,F s.t. ABDC,ABCE,and AFBC are all parallelogram choose $ G_{D} $,$ G_{E} $,and $ G_{F} $ on f(x) s.t. $ DG_{D} $,$ EG_{E} $,and $ FG_{F} $ are plumb lines now,we only show that $ \frac{y_{G_{D}}}{y_{D}}+\frac{y_{G_{E}}}{y_{E}}\leq 2 $ that is $ \frac{DG_{D}}{y_{D}}\geq \frac{EG_{E}}{y_{E}} $ it is obious because $ y_{D}\geq y_{E} $且$ DG_{D}\leq EG_{E} $

goooooooood!

Solution from Twitch Solves ISL: To deal with the long denominators, let $x = \sqrt b + \sqrt c - \sqrt a$, etc.\ for brevity. Note that $x > 0$, and similarly for the others. Then we have that \[ 0 < b+c-a = \left( \frac{x+y}{2} \right)^2 + \left( \frac{x+z}{2} \right)^2 - \left( \frac{y+z}{2} \right)^2 = x^2+xy+xz-yz \]Thus the problem asks us to prove that \[ \sum_{\text{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x} \le 3\sqrt 2 \]for positive real numbers $x$, $y$, $z$ satisfying the extra condition that all radicals are positive. By Cauchy-Schwarz though, we have that \[ \left( \sum_{\text{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x} \right)^2 \le 3 \sum_{\text{cyc}} \frac{x^2+xy+xz-yz}{x^2} \]and expanding fully, we see it suffices to show \[ \sum_{\text{sym}} xy^2z^3 \le (xy)^3 + (yz)^3 + (zx)^3 + 3(xyz)^2 \]which is Schur's inequality applied to $xy$, $yz$, $zx$.

v_Enhance wrote: Solution from Twitch stream: Then we have that \[ 0 < b+c-a = \left( \frac{x+y}{2} \right)^2 + \left( \frac{x+z}{2} \right)^2 - \left( \frac{y+z}{2} \right)^2 = x^2+xy+xz-yz \] Sir, it would be $b+c-a=\left(\frac{x+y}{2}\right)^2+\left(\frac{z+x}{2}\right)^2-\left(\frac{y+z}{2}\right)^2=\frac{2x^2+2xy+2zx-2yz}{4}=\frac{x^2+xy+xz-yz}{2}$

Solved with eisirrational, goodbear, nukelauncher, Th3Numb3rThr33. Let \(a=(q+r)^2\), \(b=(r+p)^2\), \(c=(p+q)^2\), so that the desired inequality is \[\sum_\mathrm{cyc}\frac{\sqrt{p(p+q+r)-qr}}p\le3\sqrt2.\] Then observe that \begin{align*} \left(\sum_\mathrm{cyc}\frac{\sqrt{p(p+q+r)-qr}}p\right)^2 &\le3\sum_\mathrm{cyc}\frac{p(p+q+r)-qr}{p^2}\\ &=3\left(3+\sum_\mathrm{sym}\frac pq-\sum_\mathrm{cyc}\frac{qr}{p^2}\right), \end{align*}so it suffices to show \[3+\sum_\mathrm{cyc}\frac{qr}{p^2}\ge\sum_\mathrm{sym}\frac pq.\]This is just Schur's inequality on \(p^{-2/3}q^{1/3}r^{1/3}\), \(p^{1/3}q^{-2/3}r^{1/3}\), \(p^{1/3}q^{1/3}r^{-2/3}\): \begin{align*} 3\prod\sqrt[3]{\frac{qr}{p^2}}+\sum_\mathrm{cyc}\frac{qr}{p^2} &\ge\sum_\mathrm{cyc}\sqrt[3]{\frac{qr}{p^2}}\sqrt[3]{\frac{rp}{q^2}}\left(\sqrt[3]{\frac{qr}{p^2}}+\sqrt[3]{\frac{rp}{q^2}}\right)\\ &=\sum_\mathrm{cyc}\sqrt[3]{\frac{r^2}{pq}}\left(\frac{q\sqrt[3]r+r\sqrt[3]q}{\sqrt[3]{p^2q^2}}\right)\\ &=\sum_\mathrm{cyc}\frac{qr+rp}{pq}=\sum_\mathrm{sym}\frac pq. \end{align*}

This inequality is just rationalization and schur.

Let $a=(x+y)^2$, $b=(y+z)^2$, $c=(z+x)^2$, $a$, $b$, $c\ge0$. It suffices to show that \[S=\sum_{\text{cyc}}{\frac{\sqrt{x^2+xy+xz-yz}}{x} }\le 3\sqrt{2}\]but by Schur's, $\sum_{\text{cyc}}{(xy)^2(yz+zx-xy)}\le 3x^2y^2z^2$ so \[S^2 \le (1+1+1)\left(\sum_{cyc}{\frac{z^2}{z^2}+\frac{yz+zx-xy}{z^2}}\right)\le (1+1+1)(3+3)=18\]which gives the desired result.

Let $x = \sqrt b + \sqrt c - \sqrt a$ and cyclic permutations. The inequality boils down to $$\sum \frac{\sqrt{x^2+xy+xz-yz}}x \leq 3\sqrt 2.$$But by Cauchy $$\left(\sum \frac{\sqrt{x^2+xy+xz-yz}}x\right)^2 \leq 3\sum \frac{x^2+xy+xz-yz}{x^2} \leq 18.$$The last inequality $$\sum \frac{xy+xz-yz}{x^2} \leq 3$$expands to Schur with $r=2$.

Substitute $x = \sqrt{b} + \sqrt{c} - \sqrt{a}$ and so on. This then simplifies as \[ \frac13 \sum_{\text{cyc}} \frac{\sqrt{x^2 + xz + xy - yz}}{x} \le \sqrt{2} \]By Power-Mean and Schur on $r = -2$ it follows that \[ \frac13 \sum_{\text{cyc}} \frac{\sqrt{x^2 + xz + xy - yz}}{x} \le \left(\frac13 \sum_{\text{cyc}} \frac{x^2 + xz + xy - yz}{x^2}\right)^{\frac 12} \le \left(\frac 13 (3 + 3) \right)^{\frac 12}. \]

We can first make the substitution $x = \sqrt b + \sqrt c - \sqrt a$, and cyclically for $y,z$, noting all of these variables are also positive. Then the left hand side can be expressed as \[\sum_{\text{cyc}} \sqrt{\frac{x^2+xy+xz-yz}{2x^2}} \leq \sqrt{3 \cdot \sum_{\text{cyc}} \frac{x^2+xy+xz-yz}{2x^2}}\] using Cauchy. Hence it suffices to show \[\sum_{\text{cyc}} \frac{x^2+xy+xz-yz}{x^2} \leq 6\]\[\iff (xy)^3 + (yz)^3 + (zx)^3 + 3(xy)(yz)(zx) \ge \sum_{\text{sym}} (xy)^2(yz),\] which is indeed just Schur's for $t=1$ on $xy,yz,zx$. $\blacksquare$

Make the substitution $x = \sqrt{b} + \sqrt{c} - \sqrt{a}$, with $y$ and $z$ defined cyclically. Now note, \begin{align*} \sqrt{b+c-a} = \sqrt{\frac{x^2+xy+xz-yz}{2}} \end{align*}Thus we wish to prove, \begin{align*} \sum_{cyc} \frac{\sqrt{x^2+xy+xz-yz}}{x} \leq 3\sqrt{2}\\ \left( \sum_{cyc} \frac{\sqrt{x^2+xy+xz-yz}}{x} \right)^2 \leq 18 \end{align*}However Cauchy gives, \begin{align*} \left( \sum_{cyc} \frac{\sqrt{x^2+xy+xz-yz}}{x} \right)^2 \leq 3\left(\sum_{cyc} \frac{x^2+xy+xz-yz}{x^2} \right) \end{align*}Substituting this into our original expression we find, \begin{align*} \sum_{cyc} \frac{x^2+xy+xz-yz}{x^2} &\leq 6\\ \end{align*}which is just Schurs upon expansion.

Let $x=\sqrt{b}+\sqrt{c}-\sqrt{a}>0$ (positive by the triangle inequality), etc., so $y+z=2\sqrt{a}$ and the LHS becomes $$\sum_{\mathrm{cyc}} \frac{\sqrt{(\frac{x+y}{2})^2+(\frac{x+z}{2})^2-(\frac{y+z}{2})^2}}{x}=\sum_{\mathrm{cyc}} \frac{\sqrt{x^2+xy+xz-yz}}{x\sqrt{2}}.$$It thus suffices to prove that $$\sum_{\mathrm{cyc}} \sqrt{1+\frac{y}{x}+\frac{z}{x}-\frac{yz}{x^2}} \leq 3\sqrt{2}.$$By Jensen's, it suffices to show that $$1+\frac{1}{3}\left(\sum_{\mathrm{cyc}} \frac{y}{x}+\frac{z}{x}-\frac{yz}{x^2}\right) \leq 2 \iff \sum_{\mathrm{cyc}} \frac{y}{x}+\frac{z}{x} \leq 3+\sum_{\mathrm{cyc}} \frac{yz}{x^2} \iff 3x^2y^2z^2+3x^3y^3+3y^3z^3+3z^3x^3 \geq x^3y^2z+x^3yz^2+x^2y^3z+xy^3z^2+x^2yz^3+xy^2z^3 \iff [3,3,0]+[2,2,2] \geq 2[3,2,1].$$This is just first-degree Schur on $xy,yz,zx$, so we're done. $\blacksquare$

Italian winter camp question proposed by Korea is interesting

Used the first hint on ARCH. Let $x=\sqrt b+\sqrt c-\sqrt a$. We will prove the (clearly) stronger result that the sum of the squares of these things is also $\le 3$. Note that $a=\frac14 (y+z)^2$, etc. Then we want to show \[\sum_{\text{cyc}}\frac{x^2+xy+xz-yz}{x^2}\le 6.\]But after we expand this, we realize that this is just third-degree Schur on $xy,yz,zx$. $\blacksquare$