solving $x,y,z$ in $a,b,c$ $x=\frac{b^2+c^2-a^2}{2bc},y=\frac{a^2+c^2-b^2}{2ac},z=\frac{a^2+b^2-c^2}{2ab}$ so we can assume that there is a acute triangle $ABC$ with side $a,b,c$ and $x=\cos{A},y=\cos{B},z=\cos{C}$ $\frac{x^2}{1+x}+\frac{y^2}{1+y}+\frac{z^2}{1+z}=\frac{\cos^2{A}}{\cos{A}+1}+\frac{\cos^2{B}}{\cos{B}+1}+\frac{\cos^2{C}}{\cos{C}+1}\ge \frac{1}{2}$

shohvanilu wrote: $a,b,c,x,y,z$ are positive real numbers and $bz+cy=a$, $az+cx=b$, $ay+bx=c$. Find the least value of following function $f(x,y,z)=\frac{x^2}{1+x}+\frac{y^2}{1+y}+\frac{z^2}{1+z}$ 2005 China Second Round Olympiad problem2

Solution from Twitch Solves ISL: The answer is $1/2$, achieved at $a=b=c$ and $x=y=z=1/2$. To prove the inequality, note that if we solve for $x$, $y$, $z$ in terms of $a$, $b$, $c$ we get \begin{align*} x &= \frac{-a^2+b^2+c^2}{2bc} \\ y &= \frac{a^2-b^2+c^2}{2ac} \\ z &= \frac{a^2+b^2-c^2}{2ab}. \end{align*}If $x,y,z > 0$, then it follows that $a$, $b$, $c$ are the sides of an acute triangle. Then, by the law of cosines, we actually have \[ x = \cos A, \qquad y = \cos B, \qquad z = \cos C. \]Now, let $s \coloneqq x+y+z$. Claim: We have $s \ge 3/2$. Proof. Since $0 < A, B, C < \frac{1}{2}\pi$, it follows by Jensen on $\cos$. $\blacksquare$ Then by Cauchy-Schwarz \[ \frac{x^2}{1+x}+\frac{y^2}{1+y}+\frac{z^2}{1+z} \ge \frac{(x+y+z)^2}{3+x+y+z} = \frac{s^2}{3+s} \ge \frac12 \]since $\frac{s^2}{3+s} \ge \frac{1}{2} \iff (2s-3)(s+1) \ge 0$.

v_Enhance wrote: \[ x = \cos A, \qquad y = \cos B, \qquad z = \cos C. \]Now, let $s \coloneqq x+y+z$. Claim: We have $s \ge 3/2$. Proof. Since $0 < A, B, C < \frac{1}{2}\pi$, it follows by Jensen on $\cos$. $\blacksquare$ Hm, don't you get the opposite inequality from Jensen, since cosine is concave on $\left(0, \frac{1}{2}\pi\right)$?

Oops yes. Gotta fix...

Before nearly two months ago my friend found some interesting observation around this task, by the way he was invented this task on his own, even it had been invented 8-9 years ago. This observation is that he found that x/x^2+1 +y/y^2+1 +z/z^2+1 <=6/5, also he was invented and original task(also he was solve it with (cos) and determinant), but I want to write a solution about this thing without (cos), which is good solution, by the way. Let‘s start: We have that b.z+c.y=a; a.z+c.x=b; a.y+b.x=c b.z.x+c.y.x=a.x, a.z.y+c.x.y=b.y, let take out this two, we will get b.z.x-a.z.y=a.x-b.y b.(z.x+y)=a.(z.y+x) b/a=(z.y+x)/(z.x+y) The last two: a.z.y+c.x.y=b.y; a.y.z+b.x.z=c.z, take out… c.x.y-b.x.z=b.y-c.z c.(x.y+z)=b.(x.z+y) c/b=(x.z+y)/(x.y+z) Analogically: a/c=(x.y+z)/(z.y+x) Now: (x.z+y).b/(x.y+z)=c Also: a=b.(z.x+y)/(z.y+x) From the beginning: a.z+c.x=b Therefore, after replaceable b((z.x+y).z/(z.y+x) + (x.z+y).x/(x.y+z))=b After we wave b and made this more beautiful, we wil get x^2+y^2+z^2+2.x.y.z=1 Now we will go to the real task Claim 1: x+y+z<=3/2 Proof: Let call x=a1/square root of (a1+b1).(a1+c1) And analogically, for some positive integers a1, b1, c1 The idea behind that is: after we collect this three in the equation x^2+y^2+z^2+2.x.y.z, we will get: (a1^2(c1+b1)+b1^2.(a1+c1)+c1^2(a1+b1)+2.a1.b1.c1)=? (a1+b1).(b1+c1).(a1+c1), it is clear true after we open the brackets. Now we have this interesting thing: Square root of (a1^2/(b1+a1).(a1+c1))= =square root of (a1/(b1+a1). a1/(a1+c1))<= square root of ((a1/b1+a1 +a1/c1+a1)^2/4) 4 is in the square, i.e this is equal to (a1/a1+b1+a1/a1+c1)/2 Let collect this and analogically, we will get 3/2, I.e x+y+z<=3/2 * x/x^2+1<=12.x/25 + 4/25(linearization with equation x=y=z=1/2 But we will prove that(for all fans) After we multiply this on 25.(x^2+1), we will get that(to prove that it is up to zero: 12.x^3+4.x^2-13.x+4>=0 Decomposition of this is (x-1/2).(12.x^2+10.x-8)>= But 12.x^2+10.x-8=12.(x-1/2).(x+4/3) I.e the whole thing decomposition of it is 12.(x-1/2)^2.(x+4/3), clearly it is up to 0 I.e * is true Let collect all this, we will have 12.(x+y+z)/25+12/25<=6.3/25+12/25=30/25=6/5 !!!!!by claim 1, all is perfect I hope you will like that