Inequality Proof $(x^2+1)(y^2+1)(z^2+1)\leq...$ I want to show, that for positive numbers $x,y,z$ with $xy,yz,zx\geq1$, $(x^2+1)(y^2+1)(z^2+1)\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$.
Using the AM-QM inequality for $(x^2+1),(y^2+1)$ and $(z^2+1)$, I already got $(x^2+1)(y^2+1)(z^2+1)\leq\left(\frac{x^2+y^2+z^2}{3}+1\right)^3$.By equivalent transformations of $\left(\frac{x^2+y^2+z^2}{3}+1\right)^3\leq\left(\left(\frac{x+y+z}{3}\right)^2+1\right)^3$ I get $0\leq\frac{2}{3}\left(xy+yz+zx-x^2-y^2-z^2\right)$, which cannot be true. Where did I miss something? Is the AM-QM inequality too weak? 
Thank you!
 A: Hint: First show that, if $ab\geq 1$,
$$(a^2+1)(b^2+1)\leq \left(\left(\frac{a+b}{2}\right)^2+1\right)^2$$
(hint for this: expand out and use that $(a-b)^2\geq 0$ and $ab\geq 1$). Then apply this to get that
$$(a^2+1)(b^2+1)(c^2+1)(d^2+1)\leq \left(\left(\frac{a+b+c+d}{4}\right)^2+1\right)^4$$
if $ab,bc,cd,da,ac,bd\geq 1$. Now, set $d=\frac{a+b+c}{3}$.
A: By the Carl Schildkraut's hint we have: $$(1+x^2)(1+y^2)\leq\left(1+\left(\frac{x+y}{2}\right)^2\right)^2$$ it's
$$(x-y)^2(x^2+y^2+6xy-8)\geq0,$$ which is true by AM-GM:
$$x^2+y^2+6xy-8\geq2xy+6xy-8\geq0.$$
Thus, it's enough to prove that
$$\prod_{cyc}\left(1+\left(\frac{x+y}{2}\right)^2\right)\leq\left(1+\left(\frac{x+y+z}{3}\right)^2\right)^3$$ or
$$\frac{\sum\limits_{cyc}\ln\left(1+\left(\frac{x+y}{2}\right)^2\right)}{3}\leq\ln\left(1+\left(\frac{\sum\limits_{cyc}\frac{x+y}{2}}{3}\right)^2\right),$$
which is true by Jensen because by AM-GM $$\frac{x+y}{2}\geq\sqrt{xy}\geq1$$ and for any $x\geq1$ we have:
$$\left(\ln(1+x^2)\right)''=\left(\frac{2x}{1+x^2}\right)'=\frac{2(1-x^2)}{(1+x^2)^2}\leq0.$$
A: Let $z = \max (x,y,z)$. Notice that   


*

*$z^2 \geq zx \geq 1 $ so $ z \geq 1$.

*$ (x+y+z)^2 \geq 3 (xy+yz+zx) \geq 9$ so $(x+y+z) \geq 3$.     

*Thus $z(x+y+z) \geq 3$.    

*$(x+y)^2 \geq 4xy \geq 4 $.   

*Thus $(x+y)(x+y+4z) = (x+y)^2 + 4z(x+y) \geq 4 + 8 = 12$.   


Now, onto the problem.
First, as per Carl's hint, prove that if $ xy \geq 1$, then
$$(x^2 + 1) ( y^2 + 1) \leq \left( ( \frac{x+y}{2} ) ^ 2 + 1 \right)^2.$$
Second, since $ xy \geq 1$ and $ z \left( \frac{x+y+z}{3} \right) \geq 1$, then
$$(x^2 +1)(y^2 + 1) (z^2 + 1) \left( ( \frac{x+y+z}{3} ) ^ 2 + 1 \right) \leq \left( ( \frac{x+y}{2} ) ^ 2 + 1 \right)^2 
\left( ( \frac{x+y+4z}{6} ) ^ 2 + 1 \right)^2 $$
Third, since   $\left( \frac{x+y}{2} \right) \left( \frac{x+y+4z}{6} \right) \geq 1$, then 
$$ \left( ( \frac{x+y}{2} ) ^ 2 + 1 \right)^2 \left( ( \frac{x+y+4z}{6} ) ^ 2 + 1 \right)^2 \leq \left( ( \frac{x+y+z}{3} ) ^ 2 + 1 \right)^4 $$
Hence, combining the last 2 inequalities, we get 
$$(x^2 +1)(y^2 + 1) (z^2 + 1)\leq \left( ( \frac{x+y+z}{3} ) ^ 2 + 1 \right)^3.$$

Here's the general forward-backward induction approach:   

If $ x_i x_j \geq 1$ for $ i \neq j$, then
  $$ \prod (x_i^2 + 1 ) \leq \left( \frac{ \sum x_i } { n } \right)^2 + 1) ^ n$$

Base inequality: $n = 2$. Done. 
Forward induction: $n = 2^m$.    
We induct on $m$. We have 
$$\prod_{i=1}^{2^{m+1}} (x_i^2 + 1 ) \leq \left( \frac{ \sum_{i=1}^{2^m} x_i } { 2^m } \right)^2 + 1) ^ {2^m} \left( \frac{ \sum_{i=2^m+1}^{2^{m+1}} x_i } { 2^m } \right)^2 + 1) ^ {2^m} $$
Now, $$ \left( \sum_{i=1}^{2^m} x_i \right) \left( \sum_{i=2^m+1}^{2^{m+1}} x_i \right) \geq 2^m \times 2^m$$
as the LHS consists of only cross terms, so 
$$\left( \frac{ \sum_{i=1}^{2^m} x_i } { n } \right)^2 + 1) ^ {2^m} \left( \frac{ \sum_{i=2^m+1}^{2^{m+1}} x_i } { n } \right)^2 + 1) ^ {2^m}  \leq 
\left( \frac{ \sum_{i=1}^{2^{m+1}} x_i } { n } \right)^2 + 1) ^ {2^{m+1}}.$$
This establishes the induction hypothesis.
Backward induction: $  2^m < n < 2^{m+1}$.
The issue, as Michael pointed out, is that we could have $x_1 < 1$ which could lead to $x_1 ( \frac{ \sum x_i }{n}) < 1$ which prevents us from applying the standard backward technique. We will use the above method to fix the gap.   
Let $ S  =\sum_{i=1}^n x_i$ and $ T = \sum_{i=1}^{2^m} x_i$, $K = 2^{m+1} - n$.
Standard inequality techniques show $ S \geq n $ and $ T \geq 2^m$, $S-T \geq n - 2^m$.
Define $x_{n+1} = x_{n+2} = \ldots x_{2^m} = \frac{ S} { n} \geq 1$.
Then, we have $ x_k \left( \sum_{i=2^m + 1} ^ {2^{m+1} } x_i \right) \geq 1 \times 2^m$ for $k \geq 2^m + 1$. (We use $x_1 < 1 $ here.)   
So, we can apply the inequality for the $2^m$ case, which gives us
$$ \prod_{i=1}^{2^{m+1} } ( x_i ^2 + 1) \leq \prod (\left(\frac{T}{2^m}\right)^2 + 1)^{2^m} (\left( \frac{ S-T + K \frac{S}{n} }{2^m} \right)^2 +1)^2   $$
So, to apply the base inequality again, we need to show that the product of these 2 terms is $ \geq 1$. This follows from 
$$ T ( S - T + K \frac{S}{n} ) \geq 2^m ( n-2^m + (2^{m+1}-n) \times 1 ) = 2^{m} \times 2^m$$
