Use the $pqr$ notation, so that $p+1=q+r$. Note that: \[LHS\le \sqrt{\frac{1}{3}\sum\frac{1+x^2}{1+x}}\] Note that: \[\sum\frac{1+x^2}{1+x}=p+\sum\frac{1-x}{1+x}=p+\frac{\sum (x-1)(y+1)(z+1)}{p+q+r+1}=p+\frac{-q-3r+p+3}{2p+2}=p+\frac{1-r}{p+1}=\frac{p^2+p-r+1}{p+1}=\frac{p^2+q}{p+1}\] Using $q\le \frac{p^2}{3}$, this means: \[LHS\le \sqrt{\frac{4p^2}{9(p+1)}}\] Now by $AM-GM$ we have: \[p+p+p+3\ge 4\sqrt[4]{3p^3}\implies 27(p+1)^4\ge 256p^3\implies \frac{4^4p^8}{3^8(p+1)}\le \frac{p^5}{3^5}\] Taking the $8$th root of this gives $\sqrt{\frac{4p^2}{9(p+1)}}\le RHS$, so $LHS\le RHS$ as desired.

This problem was proposed by me. The key idea was to find that $\frac{1+x^2}{1+x}=\frac{(x+y)(x+z)}{1+x+y+z}$.

What is the pqr notation? Is it related with letting x, y, z be the roots of a certain polynomial then using Vieta's? Does the fact that $x=\frac{y+z+1-yz}{yz+y+z-1}$ help?

applepi2000 wrote: Use the $pqr$ notation, so that $p+1=q+r$. Note that: Now by $AM-GM$ we have: \[p+p+p+3\ge 4\sqrt[4]{3p^3}\implies 27(p+1)^4\ge 256p^3\implies \frac{4^4p^8}{3^8(p+1)}\le \frac{p^5}{3^5}\] Taking the $8$th root of this gives $\sqrt{\frac{4p^2}{9(p+1)}}\le RHS$, so $LHS\le RHS$ as desired. Sorry for double posting! Alternatively, this could be worked backwards from the desired conclusion, so all that we need to do is to prove that $27(p+1)^4 \ge 256p^3$, which simplifies to $(p-3)^2$ times some quadratic polynomial that never hits zero

Konigsberg wrote: What is the pqr notation? It's $p=x+y+z$, $q=xy+yz+zx$, $r=xyz$. Also called $uvw$.

Sung-yoon Kim wrote: This problem was proposed by me. The key idea was to find that $\frac{1+x^2}{1+x}=\frac{(x+y)(x+z)}{1+x+y+z}$. What next?

Then use Cauchy-Schwarz to try and get a single-variable inequality in $s=x+y+z$.

Use Jensen inequality take f(x)=√((1+x²)/(1+x))

The crux of the solution is the following: Claim: (bash head against wall until claim obtained) We have $\frac{1 + z^2}{1 + z} = \frac{(x + z)(y + z)}{x + y + z + 1}$. This simplifies to the given $xyz + xy + yz + xz = x + y + z + 1$ upon expansion. $\square$ Now, observe that \begin{align*} \left( \sum \sqrt{(x + z)(y + z)} \right)^2 \leq \left(\sum x + z \right) \left(\sum y + z \right) = 4(x + y + z)^2, \end{align*}which tells us that \begin{align*} \text{LHS} \leq \frac{2(x + y + z)}{3 \sqrt{x + y + z + 1}}, \end{align*}and if we want the problem statement to hold then we need \begin{align*} \frac{2(x + y + z)}{3 \sqrt{x + y + z + 1}} &\leq \left( \frac{x + y + z}{3} \right)^{\frac 5 8} \\ \Longleftrightarrow 2^8(x + y + z)^3 &\leq 3^3(x + y + z + 1)^4, \end{align*}and we can set $x + y + z = a$ and expand both sides and try to factor. Observe that \begin{align*} 3^3(a + 1)^4 - 2^8a^3 &= 27a^4 - 148a^3 + 162a^2 + 108a + 27 \\ &= (27a^2 + 14a + 3)(a - 3)^2 \geq 0, \end{align*}and so we are done.

Here is my solution i know above are correct solution also i m just posting for storage and some standard solution so other can understand well My Solution: The main idea is here that to Claim: $\frac{(1+x^2)}{1+x}=\frac{(x+y)(x+z)}{x+y+z+1}$ this is the main idea in this problem Proof: \[\frac{(1+x^2)}{1+x}=\frac{(1+x^2)(1+y)(1+z)}{(1+x)(1+y)(1+z)}\]\begin{align*} \frac{(1+x^2)(1+y)(1+z)}{(1+x)(1+y)(1+z)}&=\frac{(1+x^2)(1+y+z+yz)}{xyz+xy+yx+xz+x+y+z+1}&\\\Leftrightarrow \frac{1+y+z+yz+x^2+x^2y+x^2z+x^2yz}{2(x+y+z+1)}&=\frac{x(xyz+xy+xz)+x^2+yz+y+z+1}{2(x+y+z+1)}&\\\Leftrightarrow \frac{x(x+y+z-yz+1)+x^2+yz+y+z+1}{2(x+y+z+1)}&=\frac{x^2+xy+xz-xyz+x+x^2+yz+y+z+1}{2(x+y+z+1)}&\\\Leftrightarrow\frac{2x^2-xyz+xy+yz+x+y+z+1}{2(x+y+z+1)}&=\frac{2x^2+2xy+2yz+2zx+2}{2(x+y+z+1)}&\\\Leftrightarrow\frac{2(x+y)(x+z)}{2(x+y+z+1)}&=\frac{(x+y)(x+z)}{x+y+z+1}\square \end{align*}So we are done here next to use this i think probabaly Cauchy's Inequality Works let's Check \[\implies\frac{\left(\sum_{cyc}(x+y)\right)\left(\sum_{cyc}(x+z)\right)}{1+x+y+z}\geq \left(\sqrt{\sum_{cyc}\frac{(x+y)(x+z)}{1+x+y+z}}\right)^2\]After some simplification we can see that \[\implies \frac{4(x+y+z)^2}{x+y+z+1}\geq \left(\sqrt{\sum_{cyc}\frac{(x+y)(x+z)}{1+x+y+z}}\right)^2 \]So its suffice to show that \[9\left(\frac{(x+y+z)}{3}\right)^\frac54 \geq \frac{4(x+y+z)^2}{x+y+z+1}\]so to do this Let $s=x+y+z$ then we can see that \[\implies 9\left(\frac{s}{3}\right)^\frac54\geq \frac{4s^2}{s+1}\]\begin{align*} \implies 9\left(\frac{s^5}{3^5}\right)\geq \frac{4^4.(s^2)^4}{(s+1)^4}&\Leftrightarrow \frac{9^4.s^5}{3^5}.(s+1)^4-(4s^2)^4\geq 0&\\\Leftrightarrow 3^3.s^5(s+1)^4-(4s^2)^4\geq 0&\Leftrightarrow3^3(s + 1)^4 - 2^8s^3&\\\Leftrightarrow27s^4 - 148s^3 + 162s^2 + 108s + 27&\implies(27a^2 + 14a + 3)(a - 3)^2 \geq 0, \end{align*}We can see that this is true for all postive $s$ So we are Done.$\blacksquare$

Here's a different (and very straightforward) proof without using $\frac{(1+x^2)}{1+x}=\frac{(x+y)(x+z)}{x+y+z+1}$ Let $x+y + z = 3k$. From $1 + 3k = \sum xy + xyz \le 3k^2 + k^3$ we get $k \ge 1$. Now $$\frac{1}{3} \left( \sqrt{\frac{1+x^2}{1+x}} + \sqrt{\frac{1+y^2}{1+y}} + \sqrt{\frac{1+z^2}{1+z}} \right)^2 \le \frac{1+x^2}{1+x} + \frac{1+y^2}{1+y} + \frac{1+z^2}{1+z} $$So it suffices to show \begin{align*} & \qquad ~~ \frac{1+x^2}{1+x} + \frac{1+y^2}{1+y} + \frac{1+z^2}{1+z} \le 3k^{\frac{5}{4}} \\ &\iff \left( \frac{1+x^2}{1+x} - x \right) + \left( \frac{1+y^2}{1+y} - y \right) + \left( \frac{1+z^2}{1+z} - z \right) \le 3k^{\frac54} - 3k \qquad \qquad (1) \end{align*}Now, \begin{align*} \text{LHS of } (1) &= \frac{1-x}{1+x} + \frac{1-y}{1+y} + \frac{1-z}{1+z} = 3 - 2 \left( \frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \right) \\ &= 3 - 2 \left( \frac{3 + 2 \sum x + \sum xy }{1 + \sum x + \sum xy + xyz} \right) = 3 - 2 \left( 1 + \frac{1 + \sum xy}{2 + 2\sum x} \right) = 1 - \frac{1 + \sum xy}{1 + 3k} \\ &= 1 - \frac{2 + 3k - xyz}{1 + 3k} = \frac{xyz - 1}{1 + 3k} \le \frac{xyz - 1}{3k} \end{align*}So we are just done if $xyz \le 1$. Otherwise suppose $xyz \ge 1$. Then, $$4(xyz)^{\frac23} \le xyz + 3(xyz)^{\frac23} \le xyz + \sum xy = 1 + 3k \le 4k\implies xyz \le k^2$$So to prove $(1)$ it suffices to show \begin{align*} \frac{k^2 - 1}{ 3k} \le 3k(k^\frac14 - 1) \iff \underbrace{ 9k^2(k^\frac14 - 1) - k^\frac32 + 1}_{ =f(k) \text{ (say} } \ge 0 \end{align*}We want $f(k) \ge 0 ~ \forall ~ k \ge 1$. As $f(1) = 0$, so it suffice to show $f'(k) \ge 0 ~ \forall ~ k \ge 1$. $$f'(k) = 18k (k^\frac14 - 1) + 9k^2 \cdot \frac{1}{4k^\frac{-3}{4}} - 2k^2 = \underbrace{18k(k^\frac14 - 1)}_{\ge 0} + \underbrace{\frac{9}{4} \cdot k^\frac{5}{4} - \frac{3}{2} k^\frac12 }_{\ge 0}$$This completes the proof. $\blacksquare$

Let $S=x+y+z$. By Jensen's Inequality, we have \begin{align*} &\frac13\left(\sqrt{\frac{1+x^2}{1+x}}+\sqrt{\frac{1+y^2}{1+y}}+\sqrt{\frac{1+z^2}{1+z}}\right)\\ \leq&\sqrt{\frac13\left(\frac{1+x^2}{1+x}+\frac{1+y^2}{1+y}+\frac{1+z^2}{1+z}\right)}\\ =&\sqrt{\frac13\frac{(1+x^2)(1+y)(1+z)+(1+y^2)(1+x)(1+z)+(1+z^2)(1+x)(1+y)}{(1+x)(1+y)(1+z)}}\\ =&\sqrt{\frac13\frac{x^2yz+xy^2z+xyz^2+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+yz+zx+xy+x^2+y^2+z^2+2x+2y+2z+3}{xyz+xy+yz+zx+x+y+z+1}}\\ =&\sqrt{\frac13\frac{xyzS+(xy+yz+zx)S-3xyz+S^2-xy-yz-zx+2S+3}{2S+2}}\\ =&\sqrt{\frac13\frac{S(S+1)+S^2+2S+3+2(xy+yz+zx)-3(S+1)}{2S+2}}\\ =&\sqrt{\frac13\frac{2S^2+2(xy+yz+zx)}{2S+2}}\\ =&\sqrt{\frac{S^2+xy+yz+zx}{3S+3}}\\ \leq&\sqrt{\frac{4S^2}{9S+9}}. \end{align*} Therefore, it suffices to show that $\left(\frac{4S^2}{9S+9}\right)^4\leq\left(\frac S3\right)^5$, or $\left(\frac S3\right)^3\leq\left(\frac{S+1}4\right)^4$. This follows from the AM-GM Inequality since $\left(\frac{S+1}4\right)^4=\left(\frac{\frac S3+\frac S3+\frac S3+1}4\right)^4\geq\left(\frac S3\right)^3$.