Tintarn wrote: Prove that for all non-negative numbers $x,y,z$ satisfying $x+y+z=1$, one has \[1 \le \frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy} \le \frac{9}{8}.\] reny inequality: $x,y,z $ are nonnegative real numbers such that $x+y+z=1$,show that \[\dfrac{x}{1-yz}+\dfrac{y}{1-zx}+\dfrac{z}{1-xy}\le\dfrac9 8\]

Tintarn wrote: Prove that for all non-negative numbers $x,y,z$ satisfying $x+y+z=1$, one has \[1 \le \frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy} \le \frac{9}{8}.\] $$\frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy} \ge x+y+z=1$$

sqing wrote: Tintarn wrote: Prove that for all non-negative numbers $x,y,z$ satisfying $x+y+z=1$, one has \[1 \le \frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy} \le \frac{9}{8}.\] $$\frac{x}{1-yz}+\frac{y}{1-zx}+\frac{z}{1-xy} \ge x+y+z=1$$ This homogenization doesn't work...

$ \sum_{cyc}\frac{x}{1-yz}=\sum_{cyc}\frac{x}{2x-x^{2}+y^{2}+yz+z^{2}}\leq \sum_{cyc}\frac{x}{2x-x^{2}+\frac{3}{4}(1-x)^{2}}$ $=4\sum_{cyc}\frac{x}{3+2x-x^{2}} \leq \frac{1}{64}\sum_{cyc}(21x+3)=\frac{9}{8} $ (toto1234567890)

For the first part $x+y+z=1\Rightarrow y+z\le 1$ thus using AM-GM $yz\le 1/4$ and we have $ \sum_{cyc}\frac{x}{1-yz}-x-y-z+1=\sum_{cyc}\frac{xyz}{1-yz} +1\ge 1$ equality holds true if at least one of the variables in $0$.

Also $\frac{x}{1 - yz} \ge \frac{x}{1} = \frac{x}{x+y+z} $ Summing up the inequality gives us the LHS ...

AdBondEvent wrote: Also $\frac{x}{1 - yz} \ge \frac{x}{1} = \frac{x}{x+y+z} $ Summing up the inequality gives us the LHS ... $\frac{x}{1 - yz} \ge x ,$ $\sum \frac{x}{1 - yz}\ge \sum x=1$ See #4

make substitution follow: \[ a=\frac{x}{1-yz}, b=\frac{y}{1-zx}, c=\frac{z}{1-xy} \]then we need proof $1 \leqslant a+b+c \leqslant 9/8$, we have \begin{align*} a & = x + ayz \\ b & = y + bzx \\ c & = z + cxy \end{align*} so we add the three equation and use $x+y+z=1$, we have \[ a+b+c=1+ayz+bzx+cxy \geqslant 1 \]

sqing wrote: $ \sum_{cyc}\frac{x}{1-yz}=\sum_{cyc}\frac{x}{2x-x^{2}+y^{2}+yz+z^{2}}\leq \sum_{cyc}\frac{x}{2x-x^{2}+\frac{3}{4}(1-x)^{2}}$ $=4\sum_{cyc}\frac{x}{3+2x-x^{2}} \leq \frac{1}{64}\sum_{cyc}(21x+3)=\frac{9}{8} $ (toto1234567890) we can write as fllow \[ \sum_{cyc}\frac{x}{1-yz} \leqslant \sum_{cyc}\frac{x}{1-(\frac{y+z}{2})^2}= \sum_{cyc}\frac{x}{1-(\frac{1-x}{2})^2} = 4\sum_{cyc}\frac{x}{3+2x-x^2} \leqslant \cdots \]

Could somebody explain me step - 4cyc(...)<=1/64cyc(...)?

$=4\sum_{cyc}\frac{x}{3+2x-x^{2}} \leq \frac{1}{64}\sum_{cyc}(63x+3)=\frac{9}{8} $

Let $a,b,c$ be non-negative numbers such that $a+b+c=1.$ Prove that $$\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca} \le \frac{27}{8}$$Let $a,b,c$ be positive numbers such that ${a^2} + {b^2} + {c^2} = 1$. Prove that: $$\frac{1}{{1 - ab}} + \frac{1}{{1 - bc}} + \frac{1}{{1 - ca}} \le \frac{9}{2}$$$$\frac{a}{1-ab}+\frac{b}{1-bc}+\frac{c}{1-ca}\le \frac{3\sqrt{3}}{2}$$

sqing wrote: $ \sum_{cyc}\frac{x}{1-yz}=\sum_{cyc}\frac{x}{2x-x^{2}+y^{2}+yz+z^{2}}\leq \sum_{cyc}\frac{x}{2x-x^{2}+\frac{3}{4}(1-x)^{2}}$ $=4\sum_{cyc}\frac{x}{3+2x-x^{2}} \leq \frac{1}{64}\sum_{cyc}(21x+3)=\frac{9}{8} $ (toto1234567890) $\frac{1}{64}\sum_{cyc}(21x+3)=\frac{9}{8} $ ,is it typo? It should be $\frac{1}{64}\sum_{cyc}(63x+3)=\frac{9}{8} $ ( Tangential Method).

OK, you're bounding x/(3+2x-x^2) above by the tangent line at x=1/3, which happens to be the optimal tangent. This seems slightly magical. How is the line chosen, and how would you use this method with sufficient explanation to gain full marks in an olympiad?

oolite wrote: OK, you're bounding x/(3+2x-x^2) above by the tangent line at x=1/3, which happens to be the optimal tangent. This seems slightly magical. How is the line chosen, and how would you use this method with sufficient explanation to gain full marks in an olympiad? It is chosen because we know that the equality happens at $x=\frac{1}{3}$ so certainly no other line could work. Now to rigorously see that it actually works, just expand the resulting inequality \[\frac{4x}{3+2x-x^2} \le \frac{63x+3}{64}\]into a polynomial \[256x \le (63x+3)(3+2x-x^2)\]and simplify to get \[(3x-1)^2(9-7x) \ge 0\]which is true (note that it's easy to find the factorization since we already know that $x=\frac{1}{3}$ should be a double root because we chose the tangent). The whole point of this method is that it reduces our three-variable problem to a one-variable inequality which we can just bash by expanding and factorizing.

@Tintarn, thanks, that's helpful. OK, so by inspection by can see x=1/3 gives the upper bound in the original problem, so that's where we should be looking to place the tangent. Got that. But how to recognise that the tangent has slope 63/256? It's easy with a little calculus, but how would a student without knowledge of calculus be able to find this? Is there a standard "recipe" that students are encouraged to learn here?

At least in Germany the students who had to solve this problem should already know the basic calculus to do this (needless to say, getting the idea is something completely different). One more elementary way to find the correct slope (which is really just an algebraic way to compute the derivative) would be to look for the value $m$ such that the equation \[\frac{4x}{3+2x-x^2}=m \cdot \left(x-\frac{1}{3}\right)+\frac{3}{8}\]has a double root at $x=\frac{1}{3}$. But the equation becomes \[m \cdot (3x-1)=\frac{96x}{3+2x-x^2}-9=\frac{96x}{3+2x-x^2}-\frac{27+18x-9x^2}{3+2x-x^2}=\frac{9x^2+78x-27}{3+2x-x^2}=\frac{(3x-1)(3x+27)}{3+2x-x^2}\]and hence we should have \[m=\frac{3x+27}{3+2x-x^2}\]at $x=\frac{1}{3}$ whence $m=\frac{63}{64}$.

The lower bound is clear by noting \[\sum_{\text{cyc}} \frac{x}{1-yz} \ge \sum_{\text{cyc}} x = 1.\] The upper bound can be handled by noting that $(y+z)^2 = (1-x)^2$ gives us \[1-yz = (y^2+yz+z^2) - (x^2-2x) \ge \frac 34(y+z)^2 - (x^2-2x) = \frac{-x^2+2x+3}{4}.\] Thus we can use the Tangent Line Trick to conclude \[\sum_{\text{cyc}} \frac{x}{1-yz} \leq \sum_{\text{cyc}} \frac{4x}{-x^2+2x+3} \leq \sum_{\text{cyc}} \left(\frac 38 + \frac{63}{64} \left(x-\frac 13\right)\right) = \frac 98.~\blacksquare\]

For clarity we shall use the dummy variables $a$, $b$, $c$ instead of $x$, $y$, $z$. The lower bound is clear as $\sum_\text{cyc}\frac{a}{1-bc}\ge a+b+c=1.$ Now, set $f(x)=\frac{4x}{3+2x-x^2}$. The line $y=\frac{63x+3}{64}$ describes the line tangent to $f$ at $x=\frac13$, and we observe that \begin{align*} \frac{63x+3}{64}&\ge\frac{4x}{3+2x-x^2}\\ \iff(3x-1)^2(7x-9)&\ge0, \end{align*}which clearly holds on $[0,1]$. It then follows by AM-GM that \[\sum_{\text{cyc}}\frac{a}{1-bc}\le\sum_{\text{cyc}}\frac{a}{1-\frac{(1-a)^2}{4}}=f(a)+f(b)+f(c)\le\frac3{64}\sum_\text{cyc}(21a+1)=\frac98,\]as desired. $\blacksquare$

We will first prove the right inequality. It is the same as proving \[\sum_{\text{cyc}} \frac{x^2}{x-xyz} \ge 1 \Longleftrightarrow \frac{(x+y+z)^2}{x+y+z-3xyz} \ge 1\]By Titu's Lemma. This simplifies to proving \[3xyz \ge 0\]which is true since $x,y,z \ge 0$. To prove the right inequality, we subtract $x+y+z$ from each side, so it suffices to prove \[\sum_{\text{cyc}} \frac{xyz}{1-yz} \le \frac{1}{8}\]\[\Longleftrightarrow \sum_{\text{cyc}} \frac{1}{1-\frac{xyz}{x}} \le \frac{1}{8xyz}\]Let $k = xyz$ be a constant. Define \[f(x) = \frac{1}{1-\frac{k}{x}} \implies f''(x) = \frac{2k}{(x-k)^3} \le 0\]Since $x < k$ since $yz<1$ since $x+y+z = 1$. Thus, applying Jensens, we get \[\frac{1}{3}\sum_{\text{cyc}} f(x) \le f\left(\frac{1}{3}\right) = \frac{1}{1-3k}\]Thus it suffices to prove that \[\frac{3}{1-3k} \le \frac{1}{8k} \Longleftrightarrow 27k \le 1\]which is true by AM-GM on $x,y,z$.