BW kills this problem!

Let $x=\frac{2a}{a-b}, y=\frac{2b}{b-c}, z=\frac{2c}{c-a}$. Then \[\left(1-\frac{2}{x}\right)\left(1-\frac{2}{y}\right)\left(1-\frac{2}{z}\right)=1,\]which is equivalent to \[xy+yz+zx-2(x+y+z)+4=0.\]If we denote $x+y+z=s$, then \begin{align*} & x^2+y^2+z^2-4(x+y+z)+k \\ =& s^2-2(xy+yz+zx)-4s+k \\ =& s^2-2(2s-4)-4s+k \\ =&(s-4)^2+k-8. \end{align*}Then $k=\boxed{8}$ is the optimal constant. Equality holds when $x+y+z=4$. One possible triple which satisfies $x+y+z=4$ and $xy+yz+zx-2(x+y+z)+4=0$ is $(x, y, z)=\left(\frac{8}{3}, \frac{2}{3}, \frac{2}{3}\right)$, which is attained for $(a, b, c)=(4, 1, -2)$.

a1267ab wrote: Let $x=\frac{2a}{a-b}, y=\frac{2b}{b-c}, z=\frac{2c}{c-a}$. Then \[\left(1-\frac{2}{x}\right)\left(1-\frac{2}{y}\right)\left(1-\frac{2}{z}\right)=1,\]which is equivalent to \[xy+yz+zx-2(x+y+z)+4=0.\]If we denote $x+y+z=s$, then \begin{align*} & x^2+y^2+z^2-4(x+y+z)+k \\ =& s^2-2(xy+yz+zx)-4s+k \\ =& s^2-2(2s-4)-4s+k \\ =&(s-4)^2+k-8. \end{align*}Then $k=\boxed{8}$ is the optimal constant. Equality holds when $x+y+z=4$. One possible triple which satisfies $x+y+z=4$ and $xy+yz+zx-2(x+y+z)+4=0$ is $(x, y, z)=\left(\frac{8}{3}, \frac{2}{3}, \frac{2}{3}\right)$, which is attained for $(a, b, c)=(4, 1, -2)$. amazing

janitor wrote: BW kills this problem! whats BW?

Buffalo way, if you interested in use search-function

achen29 wrote: janitor wrote: BW kills this problem! whats BW? Knowing what buffalo is I guess he wants to multiply all denominators.

Dadgarnia wrote: Determine the least real number $k$ such that the inequality $$\left(\frac{2a}{a-b}\right)^2+\left(\frac{2b}{b-c}\right)^2+\left(\frac{2c}{c-a}\right)^2+k \geq 4\left(\frac{2a}{a-b}+\frac{2b}{b-c}+\frac{2c}{c-a}\right)$$holds for all real numbers $a,b,c$. Proposed by Mohammad Jafari The real numbers $a,b,c $ are distinct. Prove that $$\left(\frac{a}{a-b}\right)^2+\left(\frac{b}{b-c}\right)^2+\left(\frac{c}{c-a}\right)^2\geq 2\left(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a}-1\right)$$Nice problem.

Attachments:

a1267ab wrote: Let $x=\frac{2a}{a-b}, y=\frac{2b}{b-c}, z=\frac{2c}{c-a}$. Then \[\left(1-\frac{2}{x}\right)\left(1-\frac{2}{y}\right)\left(1-\frac{2}{z}\right)=1,\]which is equivalent to \[xy+yz+zx-2(x+y+z)+4=0.\]If we denote $x+y+z=s$, then \begin{align*} & x^2+y^2+z^2-4(x+y+z)+k \\ =& s^2-2(xy+yz+zx)-4s+k \\ =& s^2-2(2s-4)-4s+k \\ =&(s-4)^2+k-8. \end{align*}Then $k=\boxed{8}$ is the optimal constant. Equality holds when $x+y+z=4$. One possible triple which satisfies $x+y+z=4$ and $xy+yz+zx-2(x+y+z)+4=0$ is $(x, y, z)=\left(\frac{8}{3}, \frac{2}{3}, \frac{2}{3}\right)$, which is attained for $(a, b, c)=(4, 1, -2)$. how did you come up with this beatiful solution

I'd like to know how many contestants solved it?

This question easily turns to imo 2008 question 2.

janitor wrote: Buffalo way, if you interested in use search-function by no means obliged to answer if you don't find my question a good/suitable/appropriate one!

achen29 wrote: janitor wrote: Buffalo way, if you interested in use search-function by no means obliged to answer if you don't find my question a good/suitable/appropriate one! Yes, of course! Solution by BW is complicated, after substitution $b=a+u$, $c=a+v$ it's quadratic polynomial of $a$, then observe when $\Delta\leq0$.

$\sum(\frac {2a}{a-b})^2+8\ge4\sum\frac {2a}{a-b}$ $\Longleftrightarrow $$\sum (\frac {2a}{a-b}-2)^2\ge4$ $\Longleftrightarrow $$\sum (\frac {b}{a-b})^2\ge1$ by Cauchy Schwarz we have: $\sum (\frac {b}{a-b})^2 (\sum (a-b)^2 (b-c)^2)\ge (\sum b (b-c))^2$ but $ \sum (a-b)^2 (b-c)^2=(\sum b (b-c))^2$ An easy way to prove this: for $x,y,z$ which $x+y+z=0$ we have: $\sum x^2y^2=(\sum xy)^2$ Now just let $x,y,z$ be $b-c,c-a,a-b$.

$$\left(\frac{2a}{a-b}\right)^2+\left(\frac{2b}{b-c}\right)^2+\left(\frac{2c}{c-a}\right)^2+8\geq 4\left(\frac{2a}{a-b}+\frac{2b}{b-c}+\frac{2c}{c-a}\right) \Longleftrightarrow \left(\frac{b}{a-b}\right)^2+\left(\frac{c}{b-c}\right)^2+\left(\frac{a}{c-a}\right)^2\ge 1$$IMO 2008, Question 2

sqing wrote: $$\left(\frac{2a}{a-b}\right)^2+\left(\frac{2b}{b-c}\right)^2+\left(\frac{2c}{c-a}\right)^2+8\geq 4\left(\frac{2a}{a-b}+\frac{2b}{b-c}+\frac{2c}{c-a}\right) \Longleftrightarrow \left(\frac{b}{a-b}\right)^2+\left(\frac{c}{b-c}\right)^2+\left(\frac{a}{c-a}\right)^2\ge 1$$ My solution uses this step. To finish it, let $a-b=x, b-c=y.$ Now, rewriting the inequality, we obtain: $$\frac{(y+c)^2}{x^2}+\frac{c^2}{y^2}+\frac{(x+y+c)^2}{(x+y)^2}-1\ge 0.$$ Now, the $LHS$ is quadratic for $c$ with a positive leading coefficient and $D=0.$ Done!

Solved with awang11, GeronimoStilton, Smileyklaws and mathgirl199. We claim that $\boxed{k=8}$. It is easy to verify that the equality can be achieved when $(a,b,c) = (4,1,-2)$. Now we show that this value of $k$ is optimal. Let $u = \frac{2a}{a-b}$, define $v,w$ similarly. Note that $(u-2)a = ub$, so $\prod_{cyc}(u-2) = \prod_{cyc} u$ or $abc = 0$. If $abc=0$ then WLOG $a = 0$, the original inequality becomes equivalent to $(\frac{b}{b-c}-1)^2 \ge 0$ which is evident. Now suppose $abc \ne 0$. Then the original inequality is equvialent to finding the smallest $k$ such that $$f(u,v,w) = u^2 + v^2 + w^2 - 4(u+v+w) + k \ge 0$$subject to $\prod u = \prod (u-2) \iff g(u,v,w):=\sum uv - 4\sum u + 8 = 0$. Note that $u^2 + v^2 + w^2 - 4u-4v-4w + 8 = \sum (u-2)^2 -4 > 0$ for all $u,v,w$ such that one of $|u|,|v|,|w| \ge 10^{100}$, so we will only consider $(u,v,w) \in [-10^{100},10^{100}]^3$ which is compact. Thus local minima of $f$ must exist. Evidently $f$ stays above $0$ at the boundary and is equal to $0$ at some point inside, it suffices to check the points $x = (u,v,w)$ such that $f'(x) = Lg'(x)$ for some real number $L$. We claim that $L = 1$. Indeed, considering partial derivatives of $f',g'$ wrt $u$ gives $u-2 = L(-v-w+2)$. Compare with the result when differentiating wrt $v$ gives $(L-1)(u-v) = 0$, thus if $L \ne 1$ then symmetrically we will have $u = v= w$, which means that $u^3 = \prod u = \prod (u-2) = (u-2)^3 \implies u = u-2$, contradiction. Therefore $L = 1$. Adding $u-w = L(-v-w+2)$ and symmetric expressions we obtain $\sum u = \frac{6(1+L)}{1+2L} = 4$, hence $g(u,v,w) = 0\implies \sum uv = 4$, so $f(x) = (\sum u)^2 - 2\sum uv - 4\sum u +k = 16 - 2\cdot 4 - 4\cdot 4 + 8 = 0$, therefore $\min f = 0$, as desired. $\blacksquare$

I’m not sure if this is equivalent to one of the above but here it goes: First an interesting algebraic identity: $$ \sum \frac{a+b}{a-b} \cdot \frac{b+c}{b-c}=-1 $$The following proof is for $k=8$ the equality cases are similar to the ones mentioned above. We return to the problem at hand, it is equivalent to: $$ \sum\left(\frac{a+b}{a-b}+1\right)^{2}+8 \geqslant 4 \sum\left(\frac{a+b}{a-b}+1\right) $$Cleaning up we get: $$ \sum\left(\frac{a+b}{a-b}\right)^{2}-2 \sum \frac{a+b}{a-b}-1 \geqslant 0 $$Finally, setting $A=\sum \frac{a+b}{a-b}$ and using the aforementioned identity $$ =\left(\sum \frac{a+b}{a-b}\right)^{2}-2 \sum \frac{a+b}{a-b}-1-2 \sum \frac{a+b}{a-b} \cdot \frac{b+c}{b-c}=A^{2}-2 A+1 \geqslant 0 $$This concludes our proof