<s>[INST] How prove this $f(x,y)=(3x+y-7)(1+x^2+xy)+9\ge 0$ Let $x,y>0$,and prove  or disprove
$$f(x,y)=(3x+y-7)(1+x^2+xy)+9\ge 0$$
I know
$$f(1,1)=(4-7)(1+1+1)+9=0$$http://www.wolframalpha.com/input/?i=%28%283x%2By-7%29%281%2Bx%5E2%2Bxy%29%2B9%29
since I have use nice methods solve follow inequality
let $a,b>0$,show that
$$9a^2(b+1)-(8a-b-1)(a+b+ab)\ge 0$$
proof:
\begin{align*}&9a^2(b+1)-(8a-b-1)(a+b+ab)\\
&=9(a^2b+a^2)-(8a^2+8ab+8a^2b-ab-b^2-ab^2-a-b-ab)\\
&=a^2b+a^2+ab^2-6ab+b^2+a+b\\
&=(a^2b-2ab+b)+(ab^2-2ab+a)+(a^2-2ab+b^2)\\
&=b(a-1)^2+a(b-1)^2+(a-b)^2\ge 0
\end{align*}
if and only if $a=b=1$
so I think my  inequality maybe can use this methods,we can 
$$(3x+y-7)(1+x^2+xy)+9=()(x-y)^2+()(x-1)^2+()(y-1)^2$$
Thank you very much!
 [/INST] Here is something strange, for which I don't really have an entirely rigorous justification, but which works and for which the calcluations seem simpler.
Firstly, (as Christian Blatter's answer does) note that for $x\to\infty$ or $y \to \infty$, we have $f(x,y)\to\infty$ as well. So $f(x, y)$ attains its minimum either on the boundary:


*

*For $x \to 0^+$, we have $f(x, y) \to (y - 7)(1) + 9 \ge (0-7)(1) + 9 = 2$.

*For $y \to 0^+$, we have $f(x, y) \to (3x - 7)(1 + x^2) + 9 \ge (-7)(1) + 9 = 2$.
or in the interior. So far, it's fine. Now here's the strange part. The minimum of $(3x-y)(1+x^2+xy)$ is the solution to
$$\begin{align}
\min \quad &1 + x^2 + xy \\
\text{s.t.} \quad & 3x + y - 7 = c\\
\end{align}
$$
for some $c$. These problems (for any $c$) can be solved by the method of Lagrage multipliers: extrema (in our case, minima) occur at points where the partial derivatives are parallel: 
$$ \left(\frac{\partial}{\partial x}(1 + x^2 + xy), \frac{\partial}{\partial y}(1 + x^2 + xy)\right) = (2x + y, x)$$
and similarly
$$ \left(\frac{\partial}{\partial x}(3x + y - 7), \frac{\partial}{\partial y}(3x + y - 7)\right) = (3, 1)$$
Setting them parallel, so that $(2x + y, x) = \lambda(3, 1)$, gives $x = y$.
Among these, $f(x, x) = (1 + 2x^2)(4x - 7) + 9 = 2(x-1)^2(4x+1)$ is minimized (for $x > 0$) at $x = 1$.
The stranger thing is that if we set the partial derivatives equal (instead of parallel) we get $x = y = 1$ directly. Is this coincidence or is there something to this?
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