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!
 A: This is a standard extremal problem on the unbounded domain $\Omega:=\{(x,y)|x\geq0, y\geq0\}$. When $x$ or $y$ get large then $f(x,y)$ gets large as well. It follows that $f$ assumes a global minimum on $\Omega$. This minimum is taken either in the interior  of $\Omega$ or on the boundary. Therefore we  have to determine the critical points of $f$, i.e., the solutions of the system
$$f_x(x,y)=0,\quad f_y(x,y)=0\ ,$$ lying  in the interior of $\Omega$. Then we have to analyze the  pullbacks
$$\phi(x):=f(x,0)=3x^3-7x^2+3x+2\qquad(x>0)$$
and
$$\psi(y):=f(0,y)=y+2\qquad(y>0)$$
of $f$ to the open boundary arcs,
and finally  we have to consider the vertex $(0,0)$.
All in all we shall obtain a "candidate list" $(x_k,y_k)_{1\leq k\leq r}$ of potential minima of $f$. If all values $f(x_k,y_k)$ $\>(1\leq k\leq r)$ are $\geq0$ then the stated inequalitiy is true. Note that in the whole process no second derivatives will be computed.
Kusavil in his answer has found the two critical points $(x_1,y_1):=(1,1)$ and $(x_2,y_2):=\bigl({1\over6},{1\over6}\bigr)$ in the interior of $\Omega$.
The condition $\phi'(x)=0$ produces the two conditionally critical points 
$$(x_3,y_3):=\left({7-\sqrt{22}\over 9},0\right),\quad (x_4,y_4):=\left({7+\sqrt{22}\over 9},0\right)$$
on the positive $x$-axis, and the condition $\psi'(y)=0$ produces no such points on the $y$-axis. Finally we put $(x_5,y_5):=(0,0)$.
Our candidate list now consists of five points. Computation shows that
$$\min_{1\leq k\leq 5} f(x_k,y_k)=\min\bigl\{0,\ {125\over54},\ 2.35958, \ 0.660995,\ 2\bigr\}=0\ .$$
It follows that the stated inequality is true.
A: 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?
