Inequality time

You can check out this solution, we have $\displaystyle\sum_{cyc}\frac{xy}{1+z}=\displaystyle\sum_{cyc}xy-\frac{xyz}{1+z}$ So we get $xy+yz+xz-xyz(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z})$ . Using these following inequalities $xy+yz+xz\leq\frac{(x+y+z)^2}{3}=\frac{1}{3}$ $xyz\leq\frac{(x+y+z)^3}{27}=\frac{1}{27}$ $-\bigg(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\bigg)\leq\frac{-9}{3+x+y+z}=\frac{-9}{4}$ We get $xy+yz+xz-xyz(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z})\leq \frac{1}{3}-\frac{1}{27}.\frac{9}{4}=\frac{1}{4}$

As $x+y+z=1$ we can rewrite the initial expression as it follows:

$\frac{yz}{2x+y+z}$ + $\frac{xz}{x+2y+z}$ + $\frac{xy}{x+y+2z}$

Now applying $\frac{ab}{a+b+c}\geq\frac{ab}{a+b+2c}$ (for $a,b,c\geq0$ ) to each term of the identity above we get:

$\frac{yz}{2x+y+z}$ + $\frac{xz}{x+2y+z}$ + $\frac{xy}{x+y+2z}\leq\frac{yz}{x+y+z}$ + $\frac{xz}{x+y+z}$ + $\frac{xy}{x+y+z}=yx+xz+xy$

On the other hand we have:

$1^2=(x+y+z)^2=x^2+y^2+z^2+2(yx+xz+xy)$

And applying AM-GM we have:

$1-2(yx+xz+xy)\geq yx+xz+xy \Leftrightarrow yx+xz+xy\leq \frac{1}{3}$

Therefore

$\frac{yz}{2x+y+z}$ + $\frac{xz}{x+2y+z}$ + $\frac{xy}{x+y+2z}\leq yx+xz+xy\leq \frac{1}{3}$

So the correct answer is, in fact, $\frac{1}{3}$ . Correct me if I'm mistaken :P

This solution is very long; however I am writing it here to pay my respects to Gurido Cuong , whose problems have taught me a lot about inequalities.The fond memories of his inequalities forced to me return to solve them once more,more than 2 years later:

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To remove the constraint,set $x=\dfrac{a}{a+b+c},y=\dfrac{b}{a+b+c},z=\dfrac{c}{a+b+c}$ , where $a,b,c$ are positive real numbers.The given expression now becomes: $\sum_{\text{cyc}}\dfrac{yz}{1+x}=\sum_{\text{cyc}}\dfrac{bc}{(a+b+c)(2a+b+c)}$

Note that the given expression is homogeneous,so WLOG set $a+b+c=3$ .The expression simplifies as follows: $\sum_{\text{cyc}}\dfrac{bc}{(a+b+c)(2a+b+c)}=\sum_{\text{cyc}}\dfrac{bc}{3(a+3)}\stackrel{\text{GM}\leq \text{AM}}\leq \sum_{\text{cyc}}\dfrac{(a-3)^2}{12(a+3)}$

Let $\dfrac{(a-3)^2}{12(a+3)}\leq \dfrac{1}{12}+k(a-1)$ for some real number $k$

$($ The value $\dfrac{1}{12}$ comes from intuition;we guess that the maximum occurs when $a=b=c=1$ which means that the L.H.S equals $\dfrac{1}{12}$ . The $(a-1)$ factor comes from the fact that we are going to add this inequalty cyclically to deduce the required maximum.When we cyclically sum it we want $k(a-m)+k(b-m)+k(c-m)=k(a+b+c-3m)=0$ ,and since we want $k\not=0$ ,we have $3m=a+b+c=3\implies m=1$ $)$

Simplifying this inequality while treating it as a quadratic in $a$ ,we get: $f(a)=(12k-1)a^2+(24k+7)a-(36k+6)\geq 0$

We get that $f^{"}(a)=2(12k-1)$ . We want $f(a)$ to be a convex function,so we set $f^{"}(a)=2(12k-1)\geq 0$ .We now see that the value $k=\dfrac{1}{12}$ will work quite nicely.(After putting it in the inequaltity and simplifying,we get (a-3)^2\leq a(a+3),which is true since \(a-3 < a and $a-3 < a+3$ )

So we have:

$\dfrac{(a-3)^2}{12(a+3)}\leq \dfrac{1}{12}+\dfrac{1}{12}(a-1)\\ \therefore \sum_{\text{cyc}}\dfrac{(a-3)^2}{12(a+3)} \leq \dfrac{3}{12}+\dfrac{1}{12}\times (a+b+c-3)=\boxed{\dfrac{1}{4}}$

Equality occurs when $a=b=c=1\implies \boxed{x=y=z=\dfrac{1}{1+1+1}=\dfrac{1}{3}}$