# 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!

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.

If we don't know what to do, we can always try to use the partial derivative test for looking for extrema of two variables function (look also @Christian Blatter answer)

$$\frac{\partial}{\partial x}f(x,y) = 9 x^2+2 x (4 y-7)+y^2-7 y+3 = 0$$ $$\frac{\partial}{\partial y}f(x,y) = 4 x^2+2xy -7x+1 = 0$$

From the second equation we have $y = \frac{7x-1-4x^2}{2x}$, so if we put it in the first one, we will get

$$-3x^2 +14x - \frac{45}{4} + \frac{1}{4x^2} = 0$$ which is equivalent to $$12x^4 - 56x^3 + 45x^2 -1 = 0$$

We already know one of the roots of this equation ($x_{1}=1$), thus by dividing polynomial above by $(x-1)$, we get

$$12x^3 - 44x^2 +x +1 =0$$

Fortunately, this equation have rational root, which we seek manually by testing $$-1,1,-\frac{1}{2},\frac{1}{2},-\frac{1}{3},\frac{1}{3},-\frac{1}{4},\frac{1}{4},-\frac{1}{6},\frac{1}{6},-\frac{1}{12},\frac{1}{12}$$ Because we only looks for $x>0$, thus we ignore negative ones, and finally get the root for $x_{2} = \frac{1}{6}$. The other two roots we get from $$(12x^3 - 44x^2 +x +1)/(x-\frac{1}{6}) = 6 (2 x^2-7 x-1) = 0$$ which are $x_{3} = \frac{1}{4} (7-\sqrt{57}) < 0$ and $x_{4} = \frac{1}{4} (7+\sqrt{57})$

Now, if we put $x_{1}, x_{2} \text{ and } x_{4}$ into $\frac{\partial}{\partial y}f(x,y) =0$, we get $$y_{1} = \frac{7x_{1}-1-4x_{1}^2}{2x_{1}} = 1$$ $$y_{2} = \frac{7x_{2}-1-4x_{2}^2}{2x_{2}} = \frac{1}{6}$$ $$y_{3} = \frac{7x_{4}-1-4x_{4}^2}{2x_{4}} = \frac{1}{4} (7-3 \sqrt{57}) < 0$$ So we can have extrema (for $x,y > 0$) in $(1,1)$ and $(\frac{1}{6},\frac{1}{6})$.

Now we seek for second partial derivatives: $$\frac{\partial^2 f(x,y)}{\partial x^2} = 18x +8y - 14$$ $$\frac{\partial^2 f(x,y)}{\partial x \partial y} = 8x +2y -7$$ $$\frac{\partial^2 f(x,y)}{\partial y^2} = 2x$$ Now, the determinant of Hessian matrix $D(x,y)$, gives us

$D(1,1) = 15 > 0$, $\frac{\partial^2 f(1,1)}{\partial x^2} = 12 > 0$, thus minimum $f(1,1) = 0$;

$D(\frac{1}{6},\frac{1}{6}) = -\frac{285}{9} < 0$, $\frac{\partial^2 f(\frac{1}{6},\frac{1}{6})}{\partial x^2} = \frac{-29}{3} < 0$, $\frac{\partial^2 f(\frac{1}{6},\frac{1}{6})}{\partial y^2} = \frac{1}{3} > 0$, thus saddle point,

$f(\frac{1}{6},\frac{1}{6}) = 9 - \frac{8664}{1296} = \frac{3000}{1296} > 0$

Also, for $x,y>0$ we have $$f(x,0) = (3x-7)(x^2+1)+9 > 0$$ $$f(0,y) = y+2 > 0$$

and I think it should be enough now. (If not, there would be extrema somewhere between, where function would need to take negative value, but except $(1,1)$ and $(\frac{1}{6},\frac{1}{6})$ there are no such points).

Sorry for so long and hope there is no formal gap above (tell me if there is, then I will edit it)

• See my completed solution. – Christian Blatter Jan 1 '14 at 10:05

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?