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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!

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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.

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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)

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